Find the cube of $(2 x + y)$

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Solution

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

Therefore,

$(2 x + y)^{3} = (2 x)^{3} + (y)^{3} + (3 \times (2 x)^{2} \times y) + (3 \times 2 x \times (y)^{2}) = 8 x^{3} + y^{3} + (3 y \times 4 x^{2}) + (6 x \times y^{2})$ $= 8 x^{3} + y^{3} + 12 x^{2} y + 6 x y^{2}$

Hence, $(2 x + y)^{3} = 8 x^{3} + y^{3} + 12 x^{2} y + 6 x y^{2}$ .

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(2 x - \sqrt{5})

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2 x - 5 y = 13

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y

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