Factorise the following: 27 y^{3}+125 z^{3} | vedclass

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(A) Using the algebraic identity $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$,we can factorise the expression.Here,$27 y^{3}+125 z^{3} = (3 y)^{3}+(5 z)^{3}$.C

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$(3y+5z)(9y^2-15yz+25z^2)$

Vedclass · Answer

(A) Using the algebraic identity $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$,we can factorise the expression.Here,$27 y^{3}+125 z^{3} = (3 y)^{3}+(5 z)^{3}$.Comparing this with $a^{3}+b^{3}$,we get $a = 3y$ and $b = 5z$.Substituting these values into the identity:$(3 y)^{3}+(5 z)^{3} = (3 y+5 z)((3 y)^{2}-(3 y)(5 z)+(5 z)^{2})$$= (3 y+5 z)(9 y^{2}-15 y z+25 z^{2})$.

Factorise the following: $27 y^{3}+125 z^{3}$

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