Class 9 RD Sharma Solutions - Chapter 4 Algebraic Identities- Exercise 4.3 | Set 2
Last Updated :
23 Jul, 2025
Algebraic identities are equations that remain true for all values for the variables involved. It is employed to simplify expressions and tackle equations. In this Exercise going to use algebraic identities to solve quadratic equations.
Solving Quadratic Equations using Algebraic Identities
A quadratic equation is any equation of ax^2 + bx + c = 0 where a, b, and c are constants. In the solution of quadratic equations, more methods are used, one of which is applying algebraic identity.
Further, we can have the following:
Method 1: Applying the Identity of (a + b)² = a² + 2ab + b²
But when the quadratic equation can be brought to form a + b = c, the coefficient of the second term, we can solve the equation by taking the square root of both sides of the equation.
Method 2: Using the Identity (a – b)2 = a2 – 2ab + b2
If the given quadratic equation is in the form of (a-b)2 = c then, there is a straightforward solution from both sides of the equation by taking the square root of it.
Method 3: Substituting the Identity a2 - b2 = (a +b)(a-b)
If the given quadratic equation can be put in the standard form a2 - b2 = 0, then we can factor the left-hand side of the equation by putting it equal to zero.
What is an Algebraic Identity?
An algebraic identity is an equation involving variables that is true for all values of those variables. Unlike true equations only for specific values of the variables, identities hold universally. These identities help simplify expressions, factorize polynomials, and solve equations more efficiently.
Chapter 4 Algebraic Identities- Exercise 4.3 | Set 2 Solution
Let's learn about, solutions for Chapter 4 Algebraic Identities- Exercise 4.3 | Set 2 in the article below:
Question 11. If 3x-2y=11 and xy = 12, find the value of 27x3-8y3
Solution:
Given, 3x-2y=11 and xy=12
we know that (a-b)3 = a3-b3-3ab(a-b)
(3x-2y)3 = 113
27x3-8y3-3(12)(11)=1331
27x3-8y3=1331+2376
27x3-8y3=3707
Hence, the value of 27x3-8y3 = 3707
Question 12. If (x4+1/x4)=119, Find the value of (x3-1/x3)
Solution:
Given, (x4+1/x4) =119 ------ 1
we know that (x+y)2 = x2+y2+2xy
substitute the given value in eq-1
(x2+1/x2)2 = x4+1/x4+2(x2)(1/x2)
= x4+1/x4+2
= 119+2
= 121
(x2+1/x2)2 = ±11
Now, find (x-1/x)
we know that (x-y)2=x2+y2-2xy
(x-1/x)2 = x2+1/x2-2×x×1/x
= x2+1/x2 -2
= 11 -2
= 9
(x-1/x) = ±3
we need to find x3-1/x3
we know that , a3-b3=(a-b)(a2+b2-ab)
x3-(1/x)3=(x-1/x)(x2+(1/x)2-x×1/x)
Here, x2+1/x2 = 11 and x-1/x=3
x3-1/x3 = 3(11+1)
= 3(12)
= 36
Hence, the value of x3-1/x3=36
Question 13. Evaluate each of the following:
(a) (103)3
(b) (98)3
(c) (9.9)3
(d) (10.4)3
(e) (598)3
(f) (99)3
Solution:
Given:
(a) (103)3
we know that (a+b)3 = a3+b3+3ab(a+b)
(103)3 can be written as (100+3)3
Here, a=100 and b=3
(103)3 = (100+3)3
=1003+33+3(100)(3)(100+3)
=1000000+27+900(103)
=1092727
The value of (103)3=1092727
(b) (98)3
we know that (a-b)3 = a3-b3-3ab(a-b)
(98)3 = (100-2)3
= 1003-23-3(100)(2)(100-2)
= 1000000-8-600(98)
= 941192
The value of (98)3 = 941192
(c) (9.9)3
we know that (a-b)3 = a3-b3-3ab(a-b)
(10-0.1)3 = (10)3-(0.1)3-3(10)(0.1)(10-0.1)
= 1000 - 0.001-3(9.9)
= 970.299
The value of (9.9)3=970.299
(d) (10.4)3
we know that (a+b)3= a3+b3+3ab(a+b)
(10+0.4)3=(10)3+(0.4)3+3(10)(0.4)(10+0.4)
= 1000+0.064+12(10.4)
= 1124.864
The value of (10.4)3=1124.864
(e) (598)3
we know that (a-b)3 = a3-b3-3ab(a-b)
(600-2)3 = (600)3-23-3(600)(2)(600-2)
= 216000000 - 8 -(3600×598)
= 216000000 -8 - 2152800
= 213847192
The value of (598)3 = 213847192
(f) (99)3
we know that (a-b)3 = a3-b3-3ab(a-b)
(100-1)3 = (100)3 -13 -3(100)(1)(100-1)
= 1000000 - 1 -300×99
= 1000000 - 1 -29700
= 970299
The value of (99)3 = 970299
Question 14. Evaluate each of the following
(a) 1113 - 893
(b) 463 +343
(c) 1043+963
(d) 933 - 1073
Solution:
Given:
(a) 1113 - 893
The above equation can be written as (100+11)3 - (100-11)3
we know that , (a+b)3 -(a-b)3= 2[b3+3a2b]
Here, a=100 b=11
(100+11)3 - (100-11)3 = 2[113+3(100)2(11)]
= 2[1331 + 330000]
= 2[331331]
= 662662
The value of 1113-893 = 662662
(b) 463 + 343
The above equation can be written as (40+6)3 - (40-6)3
we know that , (a+b)3 +(a-b)3= 2[a3+3ab2]
Here, a = 40 and b=6
(40+6)3 - (40-6)3 = 2[(40)3+3(6)2(40)]
= 2[64000+3×36×40]
=2[68320]
= 136640
The value of 463 +343=136640
(c) 1043 +963
The above equation can be written as (100+4)3 + (100-4)3
we know that, (a+b)3 +(a-b)3= 2[a3+3ab2]
here, a = 100 , b= 4
(100+4)3 +(100-4)3 = 2[(100)3 +3(100)(4)2]
= 2[1000000 + 300×16]
= 2[1004800]
= 2009600
The value of 1043 + 963 = 2009600
(d) 933 - 1073
The above equation can be written as (100-7)3 - (100+7)3
we know that, (a-b)3 -(a+b)3= -2[b3+3a2b]
here, a = 100 , b= 7
(100-7)3 - (100+7)3 = -2[73+3×(100)2×7]
= -2[210343]
= -420686
The value of 933 - 1073 = -420686
Question 15. If x+1/x = 3, calculate x2+1/x2, x3+1/x3, x4+1/x4
Solution:
Given, x+1/x=3
we know that (x+y)2= x2+y2+2xy
(x+1/x)2= x2+(1/x)2+2x(1/x)
(3)2 = x2+(1/x)2 +2
x2+1/x2 =7
squaring on both the sides,
(x2+1/x2)2 = 49
x4+1/x4+2(x2)(1/x2) = 49
x4 +1/x4 = 49-2
x4+1/x4 = 47
again cubing on both the sides,
(x+1/x)3 = x3+1/x3+3×x×1/x(x+1/x)
33 = x3+1/x3+3(3)
x3+1/x3=27-9
x3+1/x3 = 18
The value x2+1/x2 =7, x3+1/x3 = 18, x4+1/x4 = 47
Question 16. If x4+1/x4=194, calculate x2+1/x2, x3+1/x3, x+1/x
Solution:
Given,
x4+1/x4=194 ----- 1
add and subtract (2×x2×1/x2) on the left side in above given equation
x4+1/x4 +2×x2×1/x2 -2×x×1/x2 = 194
x4+1/x4+2×x2×1/x2 -2 =194
(x2)2+(1/x2)2+ 2×x2×1/x2 = 196
(x2+1/x2)2 = 196
(x2+1/x2) = 14 --------- 2
add and subtract (2×x×1/x) on the left side in above given equation
x2+1/x2+2×x×1/x-2×x×1/x =14
(x+1/x)2 = 14 +2
(x+1/x) = 4 ------------ 3
Now cubing eq-3 on both sides.
(x+1/x)3 = 43
x3+1/x3+3×x×1/x(x+1/x) = 64
x3+1/x3 +3×4 = 64
x3+1/x3 = 64 -12
= 52
Hence, the values of (x2+1/x2) = 14, (x3+1/x3) = 52, (x+1/x) = 4
Question 17. Find the value of 27x3+8y3, if
(a) 3x+2y=14 and xy = 8
(b) 3x+2y = 20 and xy=14/9
Solution:
(a) Given, 3x+2y = 14 and xy = 8
cubing on both the sides
(3x+2y)3 = 143
we know that, (a+b)3=a3+b3+3ab(a+b)
27x3+8y3+3(3x)(2y)(3x+2y) = 2744
27x3+8y3+18xy(3x+2y) = 2744
27x3+8y3+18×8×14 = 2744
27x3+8y3 = 2744 - 2016
27x3 +8y3 = 728
Hence, the value of 27x3 +8y3 = 728
(b) Given, 3x+2y = 20 and xy=14/9
cubing on both the sides
we know that, (a+b)3=a3+b3+3ab(a+b)
27x3+8y3+3(3x)(2y)(3x+2y) = 8000
27x3+8y3+18xy(3x+2y) = 8000
27x3+8y3+18×14/9×20 = 8000
27x3+8y3 = 8000 - 560
= 7440
Hence, the value of 27x3+8y3 = 7440
Question 18. Find the value of 64x3-125z3, if 4x-5z=16 and xz=12
Solution:
Given, 64x3 - 125z3
Here, 4x -5z = 16 and xz = 12
cubing (4x-5z)3 = 163
we know that (a-b)3 =a3-b3-3ab(a-b)
(4x -5z)3 = (4x)3-(5z)3-3(4x)(5z)(4x-5z)
(16)3 = 64x3 -125z3-60(4x-5z)
4096 = 64x3-125z3-60(16)
64x3-125z3 = 4096 + 960
= 5056
Hence, the value of 64x3 - 125z3 = 5056
Question 19. If x-1/x =3+2\sqrt2 , find the value of x3-1/x3
Solution:
Given, x-1/x =3+2\sqrt{2}
cubing both the sides,
we know that , (a-b)3 = a3-b3-3ab(a-b)
(x-1/x)3 = x3 -1/x3 -3×x×1/x(x-1/x)
(3+2\sqrt{2})^3 = x3-1/x3 -3(33+2\sqrt{2} )
3^3+(2\sqrt2)^3+3*3*(2\sqrt{2})(3+2\sqrt{2}) = x^3-1/x^3-3(3-2\sqrt{2})
3^3+(2\sqrt2)^3+3*3*(2\sqrt{2})(3+2\sqrt{2})+3(3-2\sqrt{2}) = x^3-1/x^3
x3-1/x3 = 108+76\sqrt{2}
Hence, the value of x3-1/x3 = 108+76\sqrt{2}
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