Expand: $(3 a + 4 b)^{3}$

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Solution

Using identity: $(x + y)^{3} = x^{3} + y^{3} + 3 x^{2} y + 3 x y^{2}$ , we get,
$(3 a + 4 b)^{3} = (3 a)^{3} + 3 (3 a)^{2} (4 b) + 3 (3 a) (4 b)^{2} + (4 b)^{3}$
$= 27 a^{3} + 108 a^{2} b + 144 a b^{2} + 64 b^{3}$
Hence, $(3 a + 4 b)^{3} = 27 a^{3} + 108 a^{2} b + 144 a b^{2} + 64 b^{3}$

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Factorise:

(3a+1/4b)³

2 x = a + \sqrt{\frac{4 b^{3} - a^{3}}{3 a}}

2 y = a - \sqrt{\frac{4 b^{3} - a^{3}}{3 a}}

x^{3} + y^{3}

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2 a + 3 d : 3 a - 4 d = 2 a^{3} + 3 b^{3} : 3 a^{3} - 4 b^{3}

{[(a + 4 b)^{3} + (a - 4 b)^{3}]}^{2}

(3 a + \frac{2}{b}) (2 a - \frac{3}{b})

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