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I can't figure out how to reduce this problem, (x^3-1)/(-3x+3)

User Nirmus
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\bf \cfrac{x^3-1}{-3x+3}\\\\ -----------------------------\\\\ 1=1^2=1^3=1^(100)=1^(1,000,000)\\\\ \textit{difference of cubes} \\ \quad \\ a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad (a+b)(a^2-ab+b^2)= a^3+b^3 \\ \quad \\ a^3-b^3 = (a-b)(a^2+ab+b^2)\qquad (a-b)(a^2+ab+b^2)= a^3-b^3\\\\ -----------------------------\\\\


\bf \cfrac{x^3-1^3}{-3x+3}\implies \cfrac{(x-1)(x^2+x+1^2)}{-3(x-1)}\implies \cfrac{x^2+x+1}{-3}
User Vincent Roye
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