Find the cube of : $5 a + 3 b$

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

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125 a^{3} + 115 a^{2} b + 75 a b^{2} + 27 b^{3}

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125 a^{3} + 135 a^{2} b + 95 a b^{2} + 27 b^{3}

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125 a^{3} + 215 a^{2} b + 65 a b^{2} + 27 b^{3}

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Solution

The correct option is A $125 a^{3} + 225 a^{2} b + 135 a b^{2} + 27 b^{3}$
$5 a + 3 b$
Cubing, we get
$= (5 a + 3 b)^{3}$
We know,
$(a + b)^{3} = a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}$
Thus,
$(5 a + 3 b)^{3} = (5 a)^{3} + (3 b)^{3} + 3 (5 a)^{2} (3 b) + 3 (5 a) (3 b)^{2}$
$= 125 a^{3} + 27 b^{3} + 225 a^{2} b + 135 a b^{2}$

Suggest Corrections

Similar questions

a + 3 b

Factorize: 125a³ – 27b³ – 225a²b + 135ab².

(5 a - 3 b) (25 a^{2} + 15 a b + 9 b^{2}) =

125 a^{3} - 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)

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