# Class 9 RD Sharma Solutions – Chapter 4 Algebraic Identities- Exercise 4.3 | Set 1

**Question 1. Find the cube of each of the following binomial **expressions**.**

**(a) (1/x+y/3)**

**(b) (3/x-2/x ^{2})**

**(c) (2x+3/x)**

**(d) (4-1/3x)**

**Solution:**

**(a) (1/x+y/3)**

Given, (1/x+y/3)

^{3}[(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)]we know that here, a = 1/x, b= y/3

By using formula:

(1/x+y/3)

^{3}= (1/x)^{3}+ (y/3)^{3}+ 3(1/x)(y/3)(1/x+y/3)= 1/x

^{3}+y^{3}/27 + y/x(1/x +y/3)= 1/x

^{3}+y^{3}/27 + y/x^{2}+ y^{2}/3xHence, (1/x+y/3)

^{3}= 1/x^{3}+y^{3}/27 + y/x^{2}+ y^{2}/3x

**(b) (3/x-2/x ^{2})**

Given, (3/x – 2/x

^{2})^{3}[(a-b)^{3}=a^{3}-b^{3}-3ab(a-b)]we know that here, a = 3/x, b= y/3

By using formula:

(3/x – 2/x

^{2})^{3}= (3/x)^{3}– (2/x^{2})^{3}-3(3/x)(2/x^{2})(3/x -2/x^{2})= 27/x

^{3 }– 8/x^{6}– 3(6/x^{3})(3/x – 2/x^{2})= 27/x

^{3}– 8/x^{6}– 18/x^{3 }(3/x – 2/x^{2})= 27/x

^{3}– 8/x^{6}– 54/x^{4}– 36/x^{5}Hence, (3/x – 2/x

^{2})^{3}= 27/x^{3}– 8/x^{6}– 54/x^{4}– 36/x^{5}

**(c) (2x+3/x)**

Given, (2x+3/x)

^{3}[(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)]we know that, a = 2x, b = 3/x

By using formula,

(2x+3/x)

^{3}= (2x)^{3}+ (3/x)^{3}+3(2x)(3/x)(2x+3/x)= 8x

^{3}+ 27/x^{3}+ 18(2x+3/x)= 8x

^{3}+ 27/x^{3}+ 36x + 54/xHence, (2x+3/x)

^{2}= 8x^{3}+ 27/x^{3}+ 36x + 54/x

**(d) (4-1/3x)**

Given, (4-1/3x)

^{3}we know that, a = 4, b = 1/3x

[(a-b)^{3}=a^{3}-b^{3}-3ab(a-b)]By using formula,

(4-1/3x)

^{3}= 4^{3}-(1/3x)^{3}-3(4)(1/3x)(4-1/3x)= 64 – (1/27x

^{3}) – 4/x(4 – 1/3x)= 64 – (1/27x

^{3}) – 16/x + 4/3xHence, (4-1/3x) = 64 – (1/27x

^{3}) – 16/x + 4/3x

**Question 2. Simplify each of the following**

**(a) (x+3) ^{3} + (x-3)^{3}**

**(b) (x/2+y/3) ^{3 }– (x/2-y/3)^{3}**

**(c) (x+2/x) ^{3 }+ (x – 2/x)^{3}**

**(d) (2x-5y) ^{3 }– (2x+5y)^{3}**

**Answer:**

(a) (x+3)^{3}+ (x-3)^{3}The above equation is in the form of a

^{3}+b^{3}=(a+b)(a^{2}+b^{2}-ab)we know that, a = (x+3), b= (x-3)

By using (a

^{3}+b^{3}) formula= (x+3+x-3)[(x+3)

^{3}+(x-3)^{3}-(x+3)(x-3)]= 2x[(x

^{2}+3^{2}+2*3*x)+(x^{2}+3^{2}-2*3*x)-(x^{2}-3^{2})]= 2x[(x

^{2}+9+6x)+(x^{2}+9-6x)-x^{2}+9]= 2x[x

^{2}+27]= 2x

^{3}+54xHence, the result of (x+3)

^{3}+(x-3)^{3}is 2x^{3}+54x

(b) (x/2+y/3)^{3}– (x/2-y/3)^{3}The above equation is in the form of a

^{3}-b^{3}=(a-b)(a^{2}+b^{2}+ab)we know that, a = (x/2+y/3), b= (x/2-y/3)

By using (a

^{3}-b^{3}) formula= [(x/2+y/3)-(x/2-y/3)][(x/2+y/3)

^{2}+(x/2-y/3)^{2}+(x/2+y/3)(x/2-y/3)]= [2y/3][((x/2)

^{2}+(y/3)^{2}+2(x/2)(y/3))+((x/2)^{2}+(y/3)^{2}-2(x/2)(y/3)+x^{2}/4-y^{2}/9]= (2y/3)[3x

^{2}/4+y^{2}/9]= x

^{2}y/2 +2y^{3}/27Hence, (x/2+y/3)

^{3}– (x/2-y/3)^{3}= x^{2}y/2 +2y^{3}/27

(c) (x+2/x)^{3}+(x-2/x)^{3}The above equation is in the form of a

^{3}+b^{3}=(a+b)(a^{2}+b^{2}-ab)we know that, a = (x+2/x), b= (x-2/x)

By using (a

^{3}+b^{3}) formula= [(x+2/x)+(x-2/x)][ (x+2/x)

^{2}+(x-2/x)^{2}– (x+2/x)(x-2/x)]= (2x)[((x)

^{2}+(2/x)^{2}+2(x)(2/x))+((x)^{2}+(2/x)^{2}-2(x)(2/x)) -(x^{2}-(2/x)^{2})]= (2x)[x

^{2}+3(2/x)^{2}]= (2x)[x

^{2}+12/x^{2}]= 2x

^{3}+ 24/xHence, (x+2/x)

^{3}+(x-2/x)^{3}= 2x^{3}+ 24/x

(d) (2x-5y)^{3}-(2x+5y)^{3}The above equation is in the form of a

^{3}-b^{3}=(a-b)(a^{2}+b^{2}+ab)we know that, a = (2x-5y), b= (2x+5y)

By using (a

^{3}-b^{3}) formula= [(2x-5y)-(2x+5y)][(2x-5y)

^{2}+(2x+5y)^{2}+(2x-5y)(2x+5y)]= (-10y)[((2x)

^{2}+(5y)^{2}-2(2x)(5y))+((2x)^{2}+(5y)^{2}+2(2x)(5y))+4x^{2}-25y^{2}]= (-10y)[3(4x

^{2})+25y^{2}]=(-10y)[12x

^{2}+25y^{2}]= -120x

^{2}y – 250y^{3}Hence, (2x-5y)

^{3}-(2x+5y)^{3}= -120x^{2}y – 250y^{3}

**Question 3. If a+b=10 and ab =21, Find the value of a**^{3}+b^{3} .

^{3}+b

^{3}.

**Solution:**

Given,

a+b = 10, ab = 21

we know that, (a+b)

^{3}= a^{3}+b^{3}+3ab(a+b) ———– 1substitute a+b = 10, ab = 21 in eq -1

â‡’ (10)

^{3}= a^{3}+b^{3}+3(21)(10)â‡’ 1000 = a

^{3}+b^{3}+630â‡’ 1000 – 630 = a

^{3}+b^{3}â‡’ 370 = a

^{3}+b^{3}Hence, a

^{3}+b^{3}=370

**Question 4. If a-b=4 and ab = 21, find the value of a**^{3}-b^{3}.

^{3}-b

^{3}.

**Solution:**

Given,

a-b = 4, ab = 21

we know that, (a-b)

^{3}=a^{3}-b^{3}-3ab(a-b)Substitute a-b=4, ab= 21 in eq-1

â‡’ (4)

^{3}= a^{3}-b^{3}-3(21)(4)â‡’ 64 = a

^{3}-b^{3}-252â‡’ 64+252 = a

^{3}– b^{3}â‡’ 316 = a

^{3}-b^{3}Hence, a

^{3}-b^{3}= 316

**Question 5. If (x+1/x) = 5, Find the value of x**^{3}+1/x^{3}

^{3}+1/x

^{3}

**Solution:**

Given, (x+1/x) = 5

we know that, (a+b)

^{3}= a^{3}+b^{3}+3ab(a+b) ————- 1Substitute (x+1/x) = 5 in eq–1

(x+1/x)

^{3}=x^{3}+(1/x)^{3}+3(x)(1/x)(x+1/x)(5)

^{3}= x^{3}+(1/x)^{3}+ 3(5)125 -15 = x

^{3}+(1/x)^{3}110 = x

^{3}+(1/x)^{3}Hence, the result is x

^{3}+1/x^{3}= 110

**Question 6. If (x-1/x) = 7, Find the value of x**^{3}-1/x^{3}

^{3}-1/x

^{3}

**Solution:**

Given, If (x-1/x) = 7

we know that, (a-b)

^{3}= a^{3}-b^{3}-3ab(a-b) ————- 1substitute (x-1/x) = 7 in eq–1

(x-1/x)

^{3}= x^{3}– 1/x^{3}-3(x)(1/x)(x-1/x)(7)

^{3}= x^{3}-1/x^{3}-3(7)343+21 = x

^{3}– 1/x^{3}364 = x

^{3}– 1/x^{3}Hence, the result is x

^{3}-1/x^{3}= 364

**Question 7. If (x-1/x) = 5, Find the value of x**^{3}-1/x^{3}

^{3}-1/x

^{3}

**Solution:**

Given, If (x-1/x) = 5

we know that, (a-b)

^{3}= a^{3}-b^{3}-3ab(a-b) ————- 1Substitute (x-1/x) = 5 in eq–1

(x-1/x)

^{3}= x^{3}– 1/x^{3}-3(x)(1/x)(x-1/x)(5)

^{3}= x^{3}-1/x^{3}-3(5)125+15 = x

^{3}– 1/x^{3}x

^{3}– 1/x^{3}=140Hence, the result is x

^{3}-1/x^{3}= 140

**Question 8. If (x**^{2}+1/x^{2}) = 51, Find the value of x^{3}-1/x^{3}.

^{2}+1/x

^{2}) = 51, Find the value of x

^{3}-1/x

^{3}.

**Solution:**

Given, If (x

^{2}+1/x^{2}) = 51we know that, (a-b)

^{2}= a^{2}+b^{2}-2ab ————- 1substitute (x

^{2}+1/x^{2}) = 51 in eq–1(x-1/x)

^{2}= x^{2}+1/x^{2}-2(x)(1/x)(x-1/x)

^{2}= 51 -2= 49

x-1/x = Â±7

we need to find x

^{3}-1/x^{3}So, a

^{3}-b^{3}= (a-b)(a^{2}+b^{2}+ab)x

^{3}-1/x^{3}= (x-1/x)(x^{2}+1/x^{2}+x*1/x)Substitute all known values here,

= 7(51+1)

=7(52)

x

^{3}-1/x^{3}= 364Hence, the result is x

^{3}-1/x^{3}= 364

**Question 9. If (x**^{2}+1/x^{2}) = 98, Find the value of x^{3}+1/x^{3}

^{2}+1/x

^{2}) = 98, Find the value of x

^{3}+1/x

^{3}

**Solution:**

Given, (x

^{2}+1/x^{2}) = 98we know that, (x+y)

^{2}= x^{2}+y^{2}+2xy ——– 1substitute (x

^{2}+1/x^{2}) = 98(x+1/x)

^{2}=x^{2}+1/x^{2}+2(x)(1/x)= 98 + 2

= 100

(x+1/x) = Â±10

We need to find x

^{3}+1/x^{3}(x+1/x) = 10 and (x

^{2}+1/x^{2}) = 98x

^{3}+1/x^{3}= 10(98-1)= 970

Hence, the value of x

^{3}+1/x^{3}= 970

**Question 10. If 2x+3y=13 and xy=6, Find the value of 8x**^{3}+27y^{3}

^{3}+27y

^{3}

**Solution:**

Given, 2x+3y=13, xy=6

We know that,

â‡’ (2x+3y)

^{3}=13^{3}â‡’ 8x

^{3}+27y^{3}+3(2x)(3y)(2x+3y) = 2197â‡’ 8x

^{3}+27y^{3}+18xy(2x+3y) = 2197â‡’ 8x

^{3}+27y^{3}+18xy(2x+3y)= 2197Substitute the values of 2x+3y=13, xy=6

â‡’ 8x

^{3}+27y^{3}+18(6)(13)= 2197â‡’ 8x

^{3}+27y^{3}= 2197 – 1404â‡’ 8x

^{3}+27y^{3}= 793Hence, the value of 8x

^{3}+27y^{3}= 793

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