# Class 9 RD Sharma Solutions – Chapter 5 Factorisation of Algebraic Expressions- Exercise 5.3

### Question 1. Factorize 64a^{3}+125b^{3}+240a^{2}b+300ab^{2}

**Solution:**

As we know that, a

^{3}+b^{3}+3a^{2}b+3ab^{2}= (a+b)^{3}Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the

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free classeswhich will definitely help them in making a wise career choice in the future.So, the above expression can be written as:

(4a)

^{3}+(5b)^{3}+3(4a)^{2}(5b)+3(4a)(5b)^{2}(4a+5b)

^{3}(4a+5b)(4a+5b)(4a+5b)

Hence, 64a

^{3}+125b^{3}+240a^{2}b+300ab^{2 }= (4a+5b)(4a+5b)(4a+5b)

### Question 2. Factorize 125x^{3}-27y^{3}-225x^{2}y+135xy^{2}

**Solution:**

As we know that, a

^{3}– b^{3}-3a^{2}b+3ab^{2}= (a-b)^{3}So, the above expression can be written as:

(5x)

^{3}-(3y)^{3}-3(5x)^{2}(3y)+3(5x)(3y)^{2 }(5x-3y)

^{3}(5x-3y)(5x-3y)(5x-3y)

Hence, 125x

^{3}-27y^{3}-225x^{2}y+135xy^{2}= (5x-3y)(5x-3y)(5x-3y)

### Question 3. Factorize 8/27x^{3}+1+4/3x^{2}+2x

**Solution:**

As we know that, a

^{3}+b^{3}+3a^{2}b+3ab^{2}= (a+b)^{3}So, the above expression can be written as:

(2/3x)

^{3}+1^{3}+3(2/3x)^{2}(1)+ 3(2/3x)(1)(2/3x+1)

^{3}(2/3x+1) (2/3x+1) (2/3x+1)

Hence, 8/27x

^{3}+1+4/3x^{2}+2x = (2/3x+1) (2/3x+1) (2/3x+1)

### Question 4. Factorize 8x^{3}+27y^{3}+36x^{2}y+54xy^{2}

**Solution:**

As we know that, a

^{3}+b^{3}+3a^{2}b+3ab^{2}= (a+b)^{3}So, the above expression can be written as:

(2x)

^{3}+(3y)^{3}+3(2x)^{2}(3y)+3(2x)(3y)^{2}(2x+3y)

^{3}(2x+3y)(2x+3y)(2x+3y)

Hence, 8x

^{3}+27y^{3}+36x^{2}y+54xy^{2}= (2x+3y)(2x+3y)(2x+3y)

### Question 5. Factorize a^{3}-3a^{2}b+3ab^{2}-b^{3}+8

**Solution:**

As we know that, a

^{3}– b^{3}-3a^{2}b+3ab^{2}= (a-b)^{3}So, the above expression can be written as:

(a-b)

^{3}+ (2)^{3}(a-b+2)[(a-b)

^{2}-(a-b)(2)+2^{2}]Use a

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

^{2}+b^{2}-2ab-2a+2b+4)Hence, a

^{3}-3a^{2}b+3ab^{2}-b^{3}+8 = (a-b+2)(a^{2}+b^{2}-2ab-2a+2b+4)

### Question 6. Factorize x^{3}+8y^{3}+6x^{2}y+12xy^{2}

**Solution:**

As we know that, a

^{3}+b^{3}+3a^{2}b+3ab^{2}= (a+b)^{3}So, the above expression can be written as:

x

^{3}+(2y)^{3}+3(x)^{2}(2y)+3(x)(2y)^{2}(x+2y)

^{3}(x+2y)(x+2y)(x+2y)

Hence, x

^{3}+8y^{3}+6x^{2}y+12xy^{2}= (x+2y)(x+2y)(x+2y)

### Question 7. Factorize 8x^{3}+y^{3}+12x^{2}y+6xy^{2}

**Solution:**

As we know that, a

^{3}+b^{3}+3a^{2}b+3ab^{2}= (a+b)^{3}So, the above expression can be written as:

(2x)

^{3}+y^{3}+3(2x)^{2}y+3(2x)(y)^{2 }(2x+y)

^{3}(2x+y)(2x+y)(2x+y)

Hence, 8x

^{3}+y^{3}+12x^{2}y+6xy^{2}= (2x+y)(2x+y)(2x+y)

### Question 8. Factorize 8a^{3}+27b^{3}+36a^{2}b+54ab^{2}

**Solution:**

As we know that, a

^{3}+b^{3}+3a^{2}b+3ab^{2}= (a+b)^{3}So, the above expression can be written as:

(2a)

^{3}+(3b)^{3}+3(2a)^{2}(3b) + 3(2a)(3b)^{2}(2a+3b)

^{3}(2a+3b) (2a+3b) (2a+3b)

Hence, 8a

^{3}+27b^{3}+36a^{2}b+54ab^{2}= (2a+3b) (2a+3b) (2a+3b)

### Question 9. Factorize 8a^{3}-27b^{3}-36a^{2}b+54ab^{2}

**Solution:**

As we know that, a

^{3}-b^{3}-3a^{2}b+3ab^{2}= (a-b)^{3}So, the above expression can be written as:

(2a)

^{3}-(3b)^{3}-3(2a)^{2}(3b) + 3(2a)(3b)^{2}(2a-3b)

^{3}(2a-3b) (2a-3b) (2a-3b)

Hence, 8a

^{3}-27b^{3}-36a^{2}b+54ab^{2}= (2a-3b) (2a-3b) (2a-3b)

### Question 10. Factorize x^{3} – 12x(x-4)-64

**Solution:**

As we know that, a

^{3}-b^{3}-3a^{2}b+3ab^{2}= (a-b)^{3}So, the above expression can be written as:

x

^{3}-12x^{2}+48x -4^{3}x

^{3}-4^{3}-3(x)^{2}(4) +3(x)(4)^{2}(x-4)

^{3}(x-4) (x-4) (x-4)

Hence, x

^{3}– 12x(x-4)-64 = (x-4) (x-4) (x-4)

### Question 11. Factorize a^{3}x^{3} -3a^{2}bx^{2} +3ab^{2}x-b^{3}

**Solution:**

As we know that, a

^{3}-b^{3}-3a^{2}b+3ab^{2}= (a-b)^{3}So, the above expression can be written as:

(ax)

^{3}-3(ax)^{2}(b)+3(ax)(b)-b^{3}(ax-b)

^{3}(ax-b) (ax-b) (ax-b)

Hence, a

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