# Freaky Factorization!

Factorize $$27x^3- 81y^3$$.

Note by Palaash Barot
2 years, 3 months ago

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$27x^3-81y^3=27(x^3-3y^3)$ You can factorize the second factor further by using the algebraic identity $$a^3-b^3=(a-b)(a^2+ab+b^2)$$, only there won't be integral coefficients in the resulting factors: $27(x^3-3y^3)=27(x-\sqrt[3]{3}y)(x^2+\sqrt[3]{3}xy+\sqrt[3]{9}y^2)$

- 2 years, 3 months ago

thanks for the help.....

- 2 years, 3 months ago