Averie Ray

2022-04-22

Simplify ${(1-x)}^{3}+2{y}^{3}-3(1-x){y}^{2}=0$ to $-(x-2y-1){(x+y-1)}^{2}=0$

impire7vw

Beginner2022-04-23Added 19 answers

Use ${a}^{3}-{b}^{3}=(a-b)({a}^{2}+ab+{b}^{2})$ and

${a}^{2}-{b}^{2}=(a-b)(a+b)$

$\begin{array}{rl}(1-x{)}^{3}+2{y}^{3}-3(1-x){y}^{2}& =(1-x{)}^{3}{-}{{y}}^{{3}}{+}{3}{{y}}^{{3}}-3(1-x){y}^{2}\\ & =(1-x{)}^{3}-{{y}}^{{3}}+3{y}^{2}(y-1+x)\\ & =(1-x-y)\left[\right(1-x{)}^{2}+(1-x)y+{y}^{2}]-3{y}^{2}(1-x-y)\\ & =(1-x-y)\left[\right(1-x{)}^{2}+(1-x)y-2{y}^{2}]\\ & =(1-x-y)\left[\right(1-x{)}^{2}-{{y}}^{{2}}+(1-x)y-{{y}}^{{2}}]\\ & =(1-x-y)\left[\right(1-x-y\left)\right(1-x+y)+y\{(1-x)-y\left\}\right]\\ & =(1-x-y{)}^{2}(1-x+2y)\end{array}$

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