Question

factorise 8x³+y³+27y³-18xyz​

Answer 1

$\red{8x^{3}+y^{3}+27y^{3}-18xyz}$

$\small{= (2x)^{3} + y^{3} + (3y)^{3} - 3\times 2x\times z \times 3y }$

$\begin{gathered} \pink { a^{3}+b^{3}+c^{3}-3abc }\\\pink {= (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca )}\end{gathered}$

$Here , a = 2x ,\:b = z \:and \: c = 3y$

$\begin{gathered} = (2x + z + 3y)[(2x)^{2}+z^{2}+(3y)^{2}-2x\times z - z \times 3y - 3y\times 2x ] \\= (2x+z+3y)(4x^{2}+z^{2}+9y^{2}-2xz-3zy-6xy) \end{gathered}$

Therefore.,

$\begin{gathered} \red{ Factors \:of \:8x^{3}+z^{3}+27y^{3}-18xyz}\\\green {= (2x+z+3y)(4x^{2}+z^{2}+9y^{2}-2xz-3yz-6xy) }\end{gathered} </p><p>$