The value of $64 a^{3} + 48 a^{2} b + 12 a b^{2} + b^{3}$ at $a = 1$ and $b = - 1$ is

25

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125

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27

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54

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Solution

The correct option is C $27$
$64 a^{3} + 48 a^{2} b + 12 a b^{2} + b^{3}$
$= {(4 a + b)}^{3}$ [ $∴$ ${(a + b)}^{3} = a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}]$
$∴$ Reqd. value $= {(4 - 1)}^{3} = 3^{3} = 27$

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Similar questions

\frac{1}{64} a^{3} - \frac{1}{16} a^{2} b + \frac{1}{12} a b^{2} - \frac{1}{27} b^{3}

(i) 8 a^{3} + b^{3} + 12 a^{2} b + 6 a b^{2} (i i) 8 a^{3} - b^{3} - 12 a^{2} b + 6 a b^{2} (i i i) 27 - 125 a^{3} - 135 a + 225 a^{2} (v i) 64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2} (v) 27 p^{3} - \frac{1}{216} - \frac{9}{2} p^{2} + \frac{1}{4} p

Factorise each of the following:

(i) (ii)

(iii) (iv)

(v)

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

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

27 - 125 a^{3} - 135 a + 225 a^{2}

64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}

27 p^{3} - \frac{1}{216} - \frac{9}{2} p^{2} + \frac{1}{4} p

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