How -x-y-3-x3-y3-3xy-x-y- can become x3-y3-x-y-3 -8203-3xy-x-y-

Dear student, #160;(x+y)3=x3+y3+3xy(x+y) #8658; #160;(x+y)3 3xy(x+y)=x3+y3 #160; #160; #160; #160; #160; #160; #160; #160; #160; #16 ...

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Addition of Algebraic Expressions

How [x+y]3=x3...

Question

How [x+y]³=x³+y³+3xy[x+y] can become x³+y³=[x+y]³-3xy[x+y]

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Solution

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x + y = 1

x^{3} + y^{3} + 3 x y = . . . . . . . . . . . .

x^{3} + y^{3} + 1 = 3 x y,

x \neq y

x + y + 1.

x + y - 1 = 0

x^{3} + y^{3} + 3 x y = 1

[\frac{x^{3} + y^{3}}{{(x - y)}^{2} + 3 x y}] \div [\frac{{(x + y)}^{2} - 3 x y}{x^{3} - y^{3}}] \times \frac{x y}{x^{2} - y^{2}}

[\frac{x^{3} + y^{3}}{(x - y)^{2} + 3 x y}] + [\frac{(x + y)^{2} - 3 x y}{x^{3} - y^{3}}] \times \frac{x y}{x^{2} - y^{2}}

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