WBBSE Chapter 5 Formulae And Their Applications
Formulae and Their Applications Introduction
In class 7 you have studied the deduction of formulae for (a + b)2, (a – b)2, and a2 – b2. In this chapter, we shall deduce the formulae for
(a + b)3, (a – b)3, a3 + b3 and a3 – b3, and by the application of these formulae we shall try to solve some algebraic problems.
Formulae of (a + b)³ and (a – b)³
1. (a + b)3=a3 + 3a2 b+ 3ab2 + b3
= a3 + b3 + 3ab (a + b).
Proof : (a + b)3 = (a + b)2 (a + b)
= (a2 + 2ab + b2) (a + b)
= a3 + a2 b+ 2a2 b+ 2ab2 + ab2 + b3
= a3 + 3 a2b + 3 ab2 + b3.
Also, (a + b)3 = a3 + 3a2b+ 3ab2 + b3
= a3 + b3 + 3 a2b + 3ab2
= a3 + b3 + 3ab (a + b).
2. (a – b)3 = a3– 3a2b+ 3ab2 – b3
= a3 – b3 – 3ab (a – b).
Proof : (a – b)3 = (a – b)2 (a – b)
= (a2 – 2a6 + 62) (a – b)
= a3 – a2b- 2a2b+ 2ab2 + ab2 – b3
= a3 – 3a2b+ 3ab2 – b3.
Also, (a – b)3 = a3 – 3a2b+ 3ab2 – b3
= a3 – b3 – 3a2 b+ 3ab2
= a3 – b3 — 3ab (a – b).
WBBSE Class 8 Formulae and Applications Notes
Geometric representation of the formula
(a + b)³ = a3 + 3a2b+ 3ab2 + b3
Read And Learn More WBBSE Solutions For Class 8 Maths
Prepare two cubes of sides 6 cm and 4 cm. Next prepare three parallelopipeds whose length and breadth are 6 cm each and height is 4 cm. Prepare three more parallelopipeds whose length and breadth are 4 cm each and height is 6 cm. Thus we get altogether eight solids.
The total volume of these eight solids = (63 + 3 x 62 x 4 + 3 x 6 x 42 + 43) cc.
Now, if we arrange these eight solids in a proper manner we get a single cube, the length of each side of which is (6 + 4) cm, and therefore its volume is
(6 + 4)3 cc.
(6 + 4)3 = 63 + 3 x 62 x 4 + 3 x 6 x 42 + 43
Now, taking 6 cm = a, 4 cm = 6,
we get, (a + b)3 = a3 + 3 a2b + Sab2 + b3.
Understanding Algebraic Formulas for Class 8
Some examples on: (a+b)3= a3+3a2b +3ab2+b3
Example 1
Find the cube of (2x + 3y).
Solution :
Given (2x + 3y).
Cube of (2x + 3y)
= (2x + 3y)3
= (2x)3+ 3 x (2x)2x 3,y + 3 x 2x (3y)2+(3y)3
= 8x² + 36x²y + 54xy2 + 27y3
Cube of (2x + 3y) = 8x² + 36x²y + 54xy2 + 27y3
Example 2
Find the cube of (ab + cd).
Solution :
Given:
(ab + cd)
Cube of (ab + cd)
= (ab + cd)3
= (ab)3 + 3 (ab)2 x cd + 3ab x (cd)2 + (cd)3
= a3b3 + 3a2b2cd + 3abc2d2 + c3d3
Cube of (ab + cd) = a3b3 + 3a2b2cd + 3abc2d2 + c3d3
Example 3
Find the cube of 101 with the help of the formula.
Solution :
Given Number 101.
Cube of 101
= (101)3 = (100 + 1)3
= (100)3 + 3 (100)2 x 1 + 3 x 100 (1)2 + (1)3
= 1000000 + 30000 + 300 + 1
= 1030301
Cube of 101 = 1030301
Common Algebraic Identities for Class 8
Example 4
Using the formula find the value of (64)3+ 3x(64)2x36 + 3x(64)x (36)2+(36)3.
Solution :
Given Formula:
(64)3+ 3x(64)2x36 + 3x(64)x (36)2+(36)3.
If we assume that 64 = a and 36 = b,
then the given expression becomes,
a3 + 3a26 + 3a62 + 63 = (a + 6)3
= (64 + 36)3 [Since a = 64, b= 36]
= (100)3
= 1000000
(64)3+ 3x(64)2x36 + 3x(64)x (36)2+(36)3 = 1000000
Example 5
With the help of the formula find the value of: 3.75 x 3.75 x 3.75 + 3 x 3.75 x 3.75 x 2.25 + 3 x 3.75 x 2.25 x 2.25 + 2.25 x 2.25 x 2.25.
Solution :
Given Formula:
3.75 x 3.75 x 3.75 + 3 x 3.75 x 3.75 x 2.25 + 3 x 3.75 x 2.25 x 2.25 + 2.25 x 2.25 x 2.25.
Assuming 3.75 = a
and 2.25 = 6 the given expression becomes,
a x a x a+3 x a x a x 6+3 x a x b x b+ b x b x b
= a3 + 3a2b+ 3ab2 + b3 = (a + b)3
= (3.75 + 2.25)3 [Since a = 3.75 and b = 2.25]
= (6.00)3
= 216
3.75 x 3.75 x 3.75 + 3 x 3.75 x 3.75 x 2.25 + 3 x 3.75 x 2.25 x 2.25 + 2.25 x 2.25 x 2.2 = 216
Example 6
If x = 2, then find the value of 27x3 + 54x2 + 36x + 8.
Solution :
Given :
x = 2
27x3 + 54x2 + 36x + 8
Substitute x = 2 In Above Equation
= (3x)3 + 3 (3x)2 x 2 + 3 (3x) x (2)2 + (2)3
= (3x + 2)3
= (3 x 2 + 2)3 [Since x = 2]
= (6 + 2)3
= (8)3
= 512
27x3 + 54x2 + 36x + 8 = 512
Example 7
If a + b – 2, then what is the value of a3 + b3 + 6ab?
Solution :
a + b = 2
Cubing both sides we get, (a + b)3 = (2)3
or, a3 + b3 + 3ab (a + b) = 8
or, a3 + 63 + 3ab x 2 = 8 [Since a + b = 2]
or, a3 + b3 + 6ab
= 8
a3 + b3 + 6ab = 8
Example 8
If xy(x + y) = 1, then what is the value x³ + y³ – 1 / x³y³?
Solution:
xy(x + y) = 1
or, x + y = 1/xy
Cubing both sides we get,
xy(x+y) = (1/xy)³
or, x3 +y3 + 3xy (x + y) = 1/x³y³
or, x3 +y +3×1=1/x³y³ [Since xy(x +y) = 1]
or, x³ + y³ +3 = 1/x³y³
or, x³ + y³ – 1/x³y³
= -3
x³ + y³ – 1 / x³y³ = -3
Example 9
If 2x + 1/3x =4, then prove that, 27x3 + 1/8x³ = 189.
Solution:
2x + 1/3x = 4
Multiplying both sides by 3/2 we get,
3x + 1/2x = 6
Cubing both sides,
(3x)³ + (1/2x)³ + 3.3x. 1/2x. (3x + 1/2x) = (6)³
or, 27×3 + 1/8×3 + 9/2 . 6 = 216
or, 27×3 + 1/8×3 27 = 216
or, 27×3 + 1/8×3 = 189 (Proved)
Example 10
If (a2 + b2)3 = (a3 + b3)2, then a/b + b/a = what?
Solution :
(a2 + b2)3 = (a3 + b3)2
or, (a2)3 + 3. (a2)2 . b2 + 3.a2. (b2)2 + (b2)3
= (a3)2 + 2a3 b3 + (b3)2
or, a3 + 3a4 b2 + 3a2 b4 + b6 = a6 + 2a3 b3 + b6
or, 3a4b2 + 3a2b4 = 2a3b3
or, 3/2 (a/b + b/a) = 1
or, a/b + b/a = 2/3
Step-by-Step Guide to Applying Algebraic Formulas
Example 11
If a + 1/ (a-5) = 7, then what is the value of (a – 5)3 + 1/ (a – 5)3
Solution:
a + 1/ (a-5) = 7
or, a – 5 + 1/ a- 5 = 2
cubing both sides we get,
\((a-5)^3+\left(\frac{1}{a-5}\right)^3+3 \cdot(a-5) \cdot \frac{1}{(\dot{a}-5)}(a-5\) \(\left.+\frac{1}{a-5}\right)=(2)^3\)
\(\text { or, }(a-5)^3+\frac{1}{(a-5)^3}+3.1 .2=8\) \(\text { or, }(a-5)^3+\frac{1}{(a-5)^3}+6=8\) \(\text { or, }(a-5)^3+\frac{1}{(a-5)^3}=2\)Example 12
If 3x + 3/x = 2, then x3 + 1/x3 + 2 = what?
Solution:
\(3 x+\frac{3}{x}=2\) \(\text { or, } x+\frac{1}{x}=\frac{2}{3}\)cubing both sides we get
\(x^3+\frac{1}{x^3}+3 \cdot x \cdot \frac{1}{x} \cdot\left(x+\frac{1}{x}\right)=\left(\frac{2}{3}\right)^3\) \(\text { or, } x^3+\frac{1}{x^3}+3 \cdot \frac{2}{3}=\frac{8}{27}\) \(\text { or, } x^3+\frac{1}{x^3}+2=\frac{8}{27}\)Example 13
Simplify : (2x + 3y)3 + 3(2x + 3y)2 (2x – 3y) + 3 (2x + 3y) (2x – 3y)2 + (2x – 3y)3.
Solution :
Let, 2x + 3y = a and 2x – 3y = b
Then the given expression = a3 + 3a2b + 3ab2 + b3
= (a + b)3 = {2x + 3y + 2x – 3y}3 [Putting a = 2x + 3y, b = 2x – 3y]
= (4x)3
= 64X3
(2x + 3y)3 + 3(2x + 3y)2 (2x – 3y) + 3 (2x + 3y) (2x – 3y)2 + (2x – 3y)3 = 64X3
Example 14
Simplify: (2a – b)3 + (2a+b)3 +12a(4a2– b2).
Solution :
The given expression = (2a – b)3 + (2a + b)3 + 12a(4a2 – b2)
= (2a – b)3 + (2a + b)3 + 3 (4a2 – b2) x 4a
= (2a- b)3 + (2a + b)3 + 3 (2a – b) (2a + b)
{(2a – b) + (2a + b)} Let, 2a – b = x
and 2a + b = y
Then the given expression = x3 + y3 + 3xy (x + y)
= (x + y)3
= (2a – b + 2a + b)3 [Putting = 2a – b, y = 2a + b]
= (4a)3
= 64a3
(2a – b)3 + (2a + b)3 + 12a(4a2 – b2) = 64a3
Practice Problems on Algebraic Formulas
Example 15
Find the cube of x + y + z.
Solution :
Cube of (x + y + z)
= (x + y + z)3
= {(x + y) + z}3
= (x + y)3 + 3(x + y)2 z + 3(x + y) (z)2 + (z)3
= x3 + 3x2y + 3xy2 + y3 + 3z (x2 + 2xy + y2) + 3 z2(x + y) + z3
= x3 + 3x²y + 3xy2 + y3 + 3x2z + 6xyz + 3yxz + 3 xz2 + 3yz2 + z3
= x3 + y3 + z3 + 3x2y + 3xy2 + 3 x2z + 3 xz2 + 3y2z + 3yz² + 6 xyz
Cube of (x + y + z) = x3 + y3 + z3 + 3x2y + 3xy2 + 3 x2z + 3 xz2 + 3y2z + 3yz² + 6 xyz
Example 16
Find the cube of 2x + 3y + 4z.
Solution :
Cube of (2x + 3y + 4z)
= (2x + 3y + 4z)3
= {(2x + 3y) + 4z}3
= (2x + 3y)3+3(2x + 3y)2x4z + 3 (2x +3y)(4 z)2 + (4 z)3
= (2x)3 + 3(2x)2 x 3y + 3 x 2x (3y)2 + (3y)3 + 12z {(2x)2 + 2 x 2x x 3y + (3y)2} + 3 x I622 (2x + 3y) + 64z3
= 8x3 + 36x2y + 54xy2 + 27y3 + 48x2z + 144xyz + 108y2z+ 96xz2 + 144yz2 + 64z3
Cube of (2x + 3y + 4z) = 8x3 + 36x2y + 54xy2 + 27y3 + 48x2z + 144xyz + 108y2z+ 96xz2 + 144yz2 + 64z3
Geometric representation of the formula (a- b)3= a3 – 3a2b + 3ab2-b3
First of all, prepare two cubes of sides 6 cm and 4 cm and also a set of three parallelopipeds of length 10 cm, breadth 6 cm, and height 4 cm.
Now, if we arrange these five solids in a proper manner we get a single cube, the length of each side of which is 10 cm. It is found that, if we remove, the three parallelopipeds of length 10 cm, breadth 6 cm, and height 4 cm and also the cube of side 4 cm from the new cube of side 10 cm then we shall be left with the cube of side 6 cm.
∴ 63 = 103 – 3 x 10 x 6 x 4 – 43.
Now since 6 = 10 – 4, the above relation gives
(10 – 4)3 = 103 – 3 x 10 (10 – 4) x 4 – 43
or, (10 – 4)3= 103 – 3x 102 x 4 + 3x 10 x 42 – 43
Now taking 10 cm = a and 4 cm = b,
we get, (a – b)3 = a3 – 3a2b + 3ab2 – b3.
Some examples on :
(a – b)3 = a3 – 3a2 b+3ab2 – b3
Example 1
Find the cube of (3a – 4b).
Solution :
Cube of (3a – 4b)
= (3a – 4b)3
= (3a)3 – 3 (3a)2 x 4b + 3 x 3a (4b)2 – (4b)3
= 27a3 – 108a2b + 144ab2 – 64b3
Cube of (3a – 4b) = 27a3 – 108a2b + 144ab2 – 64b3
Example 2
Find the cube of (pq – rs).
Solution :
Cube of (pq – rs)
= (pq – rs)3
= (PQ)3 – 3 (pq)²rs + 3pq (rs)2 – (rs)3
= p3q3 – 3p2q2rs + 3pqr2s2 – r3s3
Cube of (pq – rs) = p3q3 – 3p2q2rs + 3pqr2s2 – r3s3
Example 3
Expand : (5x – 1/5x)³
Solution :
(5x – 1/5x)³
= (5x)³ – 3. (5x)² . 1/5x + 3. 5x . (1/5x)² – (1/5x)³
= 125x³ -15x + 3/5x – 1/ 125 x³
(5x – 1/5x)³ = 125x³ -15x + 3/5x – 1/ 125 x³
Example 4
Find the cube of 99 with the help of the formula.
Solution :
Cube of 99
= (99)3
= (100 – 1)3
= (100)3-3 (100)2 x 1 + 3 x 100 x (1)2 – (1)3
= 1000000 – 30000 + 300 -1
= (1000000 + 300) – (30000 + 1)
= 1000300 – 30001
= 970299
Cube of 99 = 970299
Examples of Real-Life Applications of Algebraic Formulas
Example 5
Using the formula find the value of (75)3-3 x (75)2x 25 + 3 x 75 x (25)2 – (25)3.
Solution:
If we assume that 75 = a and 25 = b
then the given expression becomes, a3 – 3a2b + 3ab2 – b3 = (a – b)3
= (75 – 25)3 [Since a = 75, b = 25]
= (50)3
= 125000
(75)3-3 x (75)2x 25 + 3 x 75 x (25)2 – (25)3 = 125000
Example 6
With the help of the formula find the value of: 5.25 x 5.25 x 5.25 – 3 x 5.25 x 5.25 x 1.25 + 3 x 5.25 x 1.25 x 1.25 – 1.25 x 1.25 x 1.25.
Solution:
Given 5.25 x 5.25 x 5.25 – 3 x 5.25 x 5.25 x 1.25 + 3 x 5.25 x 1.25 x 1.25 – 1.25 x 1.25 x 1.25.
Assuming 5.25 = a and 1.25 = b
the given expression becomes,
= a x a x a-3 x a x a x b + 3 x a x b x b – b x b x b
= a3 – 3a2b + 3a62 – b3
= (a – b)3
= (5.25 – 1.25)3 [Since a = 5.25 and b = 1.25]
= (4.00)3
=64
5.25 x 5.25 x 5.25 – 3 x 5.25 x 5.25 x 1.25 + 3 x 5.25 x 1.25 x 1.25 – 1.25 x 1.25 x 1.25 =64
Example 7
If x – 1 and y = 2, then find the value of 125x3 – 300x²y + 240xy2 – 64y3.
Solution :
Given x = 1 And y = 2
125x3 – 300x2y + 240ry2 – 64y3
= (5x)3 – 3 (5x)2 x 4y + 3 x 5x (4y)2 – (4y)3
= (5x – 4y)3
= (5 x 1 – 4 x 2)3 [Since x = 1 and y = 2]
= (5 – 8)3
= (- 3)3
= – 27
125x3 – 300x²y + 240xy2 – 64y3 = – 27
Example 8
If x – y = 5, then what is the value of x3 – y3 – 15xy?
Solution :
Given:
x-y = 5
Cubing both sides we get, (x – y)3 = (5)3
or, x3 – y3 – 3xy (x – y) = 125
or, x3– y3 – 3xy.5 = 125
or, x3– y3 – 15xy = 125
= 125.
x3 – y3 – 15xy = 125.
Alternative method :
x3 – y3 – 15xy
= x3 – y3 – 3xy x 5
= x3 – y3 – 3xy x (x – y) [x – y = 5]
= (x- y)3
= (5)3
= 125
Example 9
If 2x – 2/x = 3, then what is the value of 8×3 – 8/x3?
Solution:
Given 2x – 2/x = 3
\(\left(2 x-\frac{2}{x}\right)^3=(3)^3\) \(\text { or, }(2 x)^3-\left(\frac{2}{x}\right)^3-3 \times 2 x \times \frac{2}{x}\left(2 x-\frac{2}{x}\right)=27\) \(\text { or, } 8 x^3-\frac{8}{x^3}-12 \times 3=27\left[\text { Since } 2 x-\frac{2}{x}=3\right]\) \(\text { or, } 8 x^3-\frac{8}{x^3}=27+36=63\)Key Terms Related to Algebraic Formulas
Example 10
Simplify : (2a + 3d)3 – 3(2a + 3b)2 (2a – 3b) + 3(2a + 3b) (2a – 3b)2– (2a – 3b)3.
Solution:
Given :-
(2a + 3d)3 – 3(2a + 3b)2 (2a – 3b) + 3(2a + 3b) (2a – 3b)2– (2a – 3b)3.
Let, 2a + 3b = x and 2a – 3b = y
Then the given expression = x3 – 3x?y + 3xy2 – y3
= (x – y)3
= {(2a + 3b) – (2a – 3b)}3 [Putting x = 2a + 3b and y – 2a – 3b]
= (2a + 3b – 2a + 3b)3
= (6b)3
= 216b3
(2a + 3d)3 – 3(2a + 3b)2 (2a – 3b) + 3(2a + 3b) (2a – 3b)2– (2a – 3b)3 = 216b3
Example 11
Simplify : (3a – 2b)3 – (2a – b)3 – 3(3a – 2b)(2a-b)(a – b).
Solution :
Given :
(3a – 2b)3 – (2a – b)3 – 3(3a – 2b)(2a-b)(a – b).
Let, 3a -2b = x and 2a – b = y
∴ x – y = (3a – 2b) – (2a – b)
= 3a – 2b – 2a + b = a – b
Then the given expression
= x3– y3 – 3xy (x – y)
= (x – y)3
= (a – b)3 [Since x – y = a – b]
= a3– 3a2 b+ 3ab2 – b3
(3a – 2b)3 – (2a – b)3 – 3(3a – 2b)(2a-b)(a – b) = a3– 3a2 b+ 3ab2 – b3
Example 12
Find the cube of a + b – c.
Solution :
Given :- a + b – c
Cube of a + b – c = (a + b – c)3 = {(a + b) – c}3
= (a + b)3 – 3 (a+b)2 c+3 (a + b) (c)2-(c)3
= a3 + 3a2b + 3ab2 + b3 – 3c (a2 + 2ab + b2) + 3c2 (a+b) – c3
= a3 + 3a26 + 3ab2 + b3 – 3a2c – 6abc – 3b2c + 3c2a + 3bc2 – c3
= a3 + 63 – c3 + 3a2 b+ 3ab2 – 3a2c + 3ac2 – 3b2c + 3bc2 – 6abc
Cube of a + b – c = a3 + 63 – c3 + 3a2 b+ 3ab2 – 3a2c + 3ac2 – 3b2c + 3bc2 – 6abc
Example 13
Find the cube of a + 2b – 3c.
Solution:
Given: a + 2b – 3c
Cube of a + 2b – 3c
= (a + 2b – 3c)3
= {(a + 2b) – 3c}3
= (a + 2b)3 – 3.(a + 2b)2. 3c + 3(a + 2b). (3c)2 – (3c)3
= a3 + 3 (a)2 x 2b + 3a (2b)2 + (2b)3 – 9c (a2 + 4ab + 4b2) + 27c2 (a + 26) – 27c3
= a3 + 6a2b + 12ab2 + 8b3 – 9a2c – 36abc – 36b2c + 27c2a+ 54bc2 – 27c3.
Cube of a + 2b – 3c = a3 + 6a2b + 12ab2 + 8b3 – 9a2c – 36abc – 36b2c + 27c2a+ 54bc2 – 27c3.
Formulae of a3+ b3 and a3 – b3
1. We have seen that,
(a + b)3 = a3 + b3 + 3ab (a + b)
Therefore, a3+b3 = (a +b)3-3ab (a+b).
2. We have seen that,
(a – b)3 = a3 – b3 — 3ab (a – b)
Therefore, a3 – b3 = (a- b)3 + 3ab (a – b)
Some Examples
Example 1
What should be multiplied with a2 – ab + b2 so that the product will be the sum of two cubes?
Solution :
(a + b) (a2 – ab + b2) = a3 + b3,
which is the sum of two cubes.
a + b
Example 2
What should be multiplied with a2 + ab + b2, so that the product will be the difference between two cubes?
Solution :
(a – b)(a2 + ab + b2) = a3 – b3,
which is the difference of two cubes.
= a – b
Conceptual Questions on Algebraic Applications
Example 3
If the sum of the cubes of the two quantities is equal to the cube of their sum, then either at least one of them will be equal to zero or their sum will be zero.
Solution :
Let the two quantities be x and y x3 + y3 = (x + y)3
or, x3 + y3 = x3 + y3 + 3xy(x + y)
or, 3xy(x + y) = 0
or, xy(x + y) = 0
∴ Either r = 0 or, y = 0
or x+y = 0 Hence proved.
Example 4
If a + b = 5 and ab = 4, then what is the value of a3 + b3?
Solution :
Given a + b = 5 And ab = 4
a3 + b3 = (a + b)3 – 3ab (a + b)
= (5)3 – 3 x 4 x 5 [Since a + b = 5 and ab = 4]
= 125 – 60
= 65
a3 + b3 = 65
Example 5
If P3 + q3 – 152 and p + q = 8, then what is the value of pq?
Solution :
Given P3 + q3 – 152 And p + q = 8.
We know that,
(p + q)3 = p3 + q3 + 3pq (p + q)
or, (8)3 = 152 + 3pq x 8
or, 512 = 152 + 24pq
or, pq = 360/24
= 15
Example 6
If x + 1/x = √3, then what is value of x³ + 1/x³
Solution:
Given x + 1/x = √3.
x³ + 1/x³ = (x + 1/x)³ – 3x 1/x . ( x + 1/x)
= (√3)³ – 3√3
= 3√3 – 3√3
= 0
x³ + 1/x³ = 0
Example 7
If 3x + 1/2x = 5, then what is the value of 27x³ + 1/8x³?
Solution:
Given 3x + 1/2x = 5.
27x³ + 1/8x³ = (3x)³ + (1/2x)³
= (3x + 1/2x)³ – 3 x 3x 1/2x .(3x +1/2x)
= (5)³ – 9/2 x 5
= 125 – 45/2
= 250 – 45 / 2
= 205/2
= 102 ½
27x³ + 1/8x³ = 102 ½
Example 8
Find the product of (3x + 4y) and (9x² – I2xy + 16y2) with the help of the formula.
Solution :
Given (3x + 4y) And (9x² – I2xy + 16y2)
(3x + 4y) (9X2 – 12xy + 16y2)
= (3x + 4y) {(3x)2 – 3x x 4y + (4y)2}
= (3x)3 + (4y)3 = 27x3 + 64y3
The required product = 27x3 + 64y3.
Example 9
Find the product of (2x – y) and (4x2 + 2xy + y2) with the help of the formula.
Solution :
Given
(2x – y) And (4x2 + 2xy + y2)
(2x – y) (4X2 + 2xy + y2)
= (2x – y) {(2x)2 + 2xy + (y)2}
= (2x)3 – (y)3
= 8x3 – y3
The required product = 8x3 – y3.
Example 10
If a – b = 5 and ab = 3, then what is the value of a3 – b3?
Solution :
Given
a – b = 5
ab = 3
a3 – b3 = (a – b)3 + 3ab (a – b)
= (5)3 + 3 x 3 x 5
= 125 + 45
= 170
a3 – b3 = 170
Example 11
If a – b = 1 and a3 – b3 = 61, then what is the value of ab?
Solution :
Given a – b = 1 And a3 – b3 = 61.
(a – b)3 = a3 – b3 – 3ab(a – b)
or, (1)3 = 61 – 3ab x 1
or, 3ab = 61 – 1 = 60
or, ab = 60/3
= 20
ab = 20
Example 12
If 2x – 1/3x, then what is the value of 8x³ – 1/27x³?
Solution:
Given 2x – 1/3x
8x³ – 1/27x³ = (2x)³ – (1/3x)³
= (2x – 1/3x) + 3 x 2x x 1/3x (2x – 1/3x)
= (3)³ + 2 x 3 [Since 2x -1/3x = 3]
= 27 + 6
= 33
8x³ – 1/27x³ = 33
Example 13
If 3a – 3/a + 1 = 0, what is the value of a³ – 1/a³ + 2?
Solution:
Given, 3a – 3/a + 1 = 0 And Given Equation a³ – 1/a³ + 2
or, 3(a – 1/a) = – 1
or, (a- 1/a) = – 1/3
Now, a³ – 1/a³ + 2
= (a – 1/a) + 3.a.1/a(a – 1/a) + 2
= (-1/3)³ + 3.(-1/3) + 2 [a – 1/a = – 1/3]
= – 1/27 – 1 + 2
= 1 – 1/27
= 26/27
a³ – 1/a³ + 2 = 26/27
Example 15
If p + 1/ p+2 = 1, then what is the value of (p+2)³ + 1/(p+2)³
Solution:
Given \(p+\frac{1}{p+2}=1\)
or, \(p+\frac{1}{p+2}=1\)
Now , \((p+2)^3+\frac{1}{(p+2)^3}\)
\(=\left[(p+2)+\frac{1}{(p+2)}\right]^3-3 \cdot(p+2) \cdot \frac{1}{(p+2)}\) \(\left[(p+2)+\frac{1}{(p+2)}\right]\) \(=[p+2+1-p]^3-3 \cdot[p+2+1-p]\) \(\left[\frac{1}{p+2}=1-p\right]\) \(=3^3-3.3=27-9=18\)(p+2)³ + 1/(p+2)³ = 18