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QUESTION: 1

Expansion and simplification of 8(3h - 4) + 5(h - 2) gives

Solution:

Now, 8 (3h - 4) + 5 (h - 2)

= 24h - 32 + 5h - 10, the expansion

= 24h + 5h - 32 - 10

= 29h - 42, the simplification

QUESTION: 2

Evaluate (11)^{3}

Solution:

11³= (10+1)³=1000+1+30(11)=1001+330=1331

QUESTION: 3

Factorise:(3x- 5y)^{3}+ (5y – 2z)^{3} + (2z – 3x)^{3}

Solution:

We don't have to cube them all to factorise. Now, One thing should be remembered that, in time of this type of factorisation we have to first add the polynomials to check if the sum is zero or not.If it is zero, then the factorisation will be in form of 3(abc). Now, lets sum them, (3x-5y) +(5y-2z) +(2z-3x).The answer is zero. So put the expression in form of 3(abc). So the factorisation of the expression is 3(3x-5y)(5y-2z)(2z-3x).

QUESTION: 4

Evaluate: 5^{3} – 2^{3} – 3^{3}

Solution:

5^{3} - 2^{3} - 3^{3}

= 125 - 8 - 27

= 90

QUESTION: 5

Factorise : 8a^{3}+ b^{3} + 12a^{2}b + 6a b^{2}

Solution:

8a^{3} + b^{3}+ 12a^{2}b + 6ab^{2}

8a^{3} + b^{3}+ 12a^{2}b + 6ab^{2}

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

= (2a + b)^{3} | Using Identity VI

= (2a + b)(2a + b)(2a + b)

QUESTION: 6

Evaluate (x+y+z)^{2}

Solution:

QUESTION: 7

Factorise : 8a^{3} + b^{3} + 12a^{2} b + 6ab^{2}

Solution:

8a^{3} + b^{3}+ 12a^{2}b + 6ab^{2}

8a^{3} + b^{3}+ 12a^{2}b + 6ab^{2}

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

= (2a + b)^{3} | Using Identity VI

= (2a + b)(2a + b)(2a + b)

QUESTION: 8

p^{3}+ q^{3} + r^{3} – 3pqr = ?

Solution:

QUESTION: 9

Factorize: 125a^{3} – 27b^{3} – 225a^{2}b + 135ab^{2}.

Solution:

QUESTION: 10

Evaluate (104)^{3}

Solution:

(104)³

=(100 + 4)³

Use ( a +b)³ = a³ + b³ + 3ab( a + b)

then,

( 100 + 4)³ = ( 100)³ +(4)³ +3(100)(4)(100+4)

=(10²)³ + 64 + 1200(104)

=1000000 + 64 + 124800

=1124864.

QUESTION: 11

Factorize: (x – y)^{3} + (y – z)^{3} + (z – x)^{3}

Solution:

QUESTION: 12

Factorise: x^{7}y + xy^{7}

Solution:

QUESTION: 13

What is the value 8^{3} – 3^{3} (without solving the cubes)?

Solution:

(8-3)((8)² + 8×3 + (3)²) = 5(64 + 24 + 9) = 5×97 = 485

QUESTION: 14

p^{3}+ q^{3} = ?

Solution:

QUESTION: 15

What is the value of 5^{3} – 1^{3}(without solving cube)?

Solution:

5^{3} - 1^{3 }can be solved using the identity ;

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

5^{3} - 1^{3} = ( 5 - 1 ) ( 5^{2} + 1^{2} + 5*1)

=(4) (25 + 1 + 5)

= (4) (31)

= 124

QUESTION: 16

Factorize: 27x^{3} – 125y^{3}

Solution:

QUESTION: 17

a^{3} – b^{3} equals

Solution:

It is the identity equation

In the fraction equation

QUESTION: 18

Evaluate: 30^{3} + 20^{3} – 50^{3}.

Solution:

We have to find the value of

QUESTION: 19

If , the value of x^{3} – y^{3} is :

Solution:

QUESTION: 20

Factorization of 103²-9 yields

Solution:

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