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$(x + y)^{3} =$

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Solution

We have,

${(x + y)}^{3}$

$= (x + y) {(x + y)}^{2}$

$= (x + y) (x^{2} + y^{2} + 2 x y)$

$= x^{3} + x y^{2} + 2 x^{2} y + x^{2} y + y^{3} + 2 x y^{2}$

$= x^{3} + 3 x y^{2} + 3 x^{2} y + y^{3}$

Hence, this is the answer.

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If x+y=4 & xy=45 then find the value of x³-y³?

Solve this fast📝

\frac{x^{3} + y^{3}}{(x + y)^{2}} + \frac{x^{3} - y^{3}}{(x - y)^{2}} = \frac{43 x}{8}, 5 x - 7 y = 4

\frac{x^{3} - y^{3}}{x - y}

x = 3

y = 2

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