Factorise (x/7+y/3)(x²/49+y²/9-xy/21)
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they are already in factor
they falls in the formula of a^3 + b^3 = (a+b)(a^2 - ab + b^2)
your answer in unfactorise form is
(X/7)^3 + (y/9)^3
they falls in the formula of a^3 + b^3 = (a+b)(a^2 - ab + b^2)
your answer in unfactorise form is
(X/7)^3 + (y/9)^3
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