Find the cube of $(4 a + 5 b)$

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Solution

Using the identity: $(a + b)^{3} = a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}$

Therefore,

$(4 a + 5 b)^{3} = (4 a)^{3} + (5 b)^{3} + (3 \times (4 a)^{2} \times 5 b) + (3 \times 4 a \times (5 b)^{2}) = 64 a^{3} + 125 b^{3} + (15 b \times 16 a^{2}) + (12 a \times 25 b^{2})$
$= 64 a^{3} + 125 b^{3} + 240 a^{2} b + 300 a b^{2}$

Hence, $(4 a + 5 b)^{3} = 64 a^{3} + 125 b^{3} + 240 a^{2} b + 300 a b^{2}$ .

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