Question

Find the cube of $(2x-\sqrt{5})$

Answer 1

We know that the identity is $(a-b{)}^3=a^3-b^3-3a^{2}b+3a{b}^2$

Therefore,

${(2x-\sqrt{5})}^3=(2x{)}^3-{\left(\sqrt{5}\right)}^3-(3\times (2x{)}^2\times \sqrt{5})+(3\times 2x\times {\left(\sqrt{5}\right)}^2)=8x^3-5\sqrt{5}-(3\sqrt{5}\times 4x^2)+(6x\times 5)$

$=8x^3-5\sqrt{5}-12\sqrt{5}{x}^2+30x$

Hence, ${(2x-\sqrt{5})}^3=8x^3-5\sqrt{5}-12\sqrt{5}{x}^2+30x$.