\[27x^3-81y^3=27(x^3-3y^3)\]
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:
\[27(x^3-3y^3)=27(x-\sqrt[3]{3}y)(x^2+\sqrt[3]{3}xy+\sqrt[3]{9}y^2)\]

## Comments

Sort by:

TopNewest\[27x^3-81y^3=27(x^3-3y^3)\] 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: \[27(x^3-3y^3)=27(x-\sqrt[3]{3}y)(x^2+\sqrt[3]{3}xy+\sqrt[3]{9}y^2)\]

Log in to reply

thanks for the help.....

Log in to reply