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8x^3-27y^3
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so what is the question ? factorize ?
Yea factor the polynomial completely
no i don't think so
No, this can be factored! It's a difference of cubes. \[a^3 - b^3 = (a-b)(a^2+ab + b^2)\]
\[a^3 = 8x^3\]\[a = \sqrt[3]{8}*\sqrt[3]{x^3} = 2x\]\[b^3=27y^3 \]\[b = \sqrt[3]{27}\sqrt[3]{y^3} = 3y\] Now fill in the template to get the factoring...
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If you haven't seen the difference of cubes formula before: \[(a-b)(a^2+ab+b^2) = a(a^2+ab+b^2) - b(a^2+ab+b^2)\]\[\qquad = a^3 + a^2b + ab^2 -a^2b - ab^2-b^3\]\[\qquad=a^3 - b^3\]
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