Question
Find the cube of the following binomial expressions: $2\text{x}+\frac{3}{\text{x}}$
✓
Answer
Given, $2\text{x}+\frac{3}{\text{x}}$ The above equation is in the form of $(a + b)^3= a^3 + b^3 + 3ab(a + b)$
we know that $\text{a}=2\text{x},\text{b}=\frac{3}{\text{x}}$ By using $(a + b)^3$
formula $=8\text{x}^3+\frac{27}{\text{x}^3}+\frac{18\text{x}}{\text{x}}\Big(\frac{2}{\text{x}}+\frac{3}{\text{x}}\Big)$
$=8\text{x}^3+\frac{27}{\text{x}^3}+\frac{18\text{x}}{\text{x}}\Big(2{\text{x}}+\frac{3}{\text{x}}\Big)$
$=8\text{x}^3+\frac{27}{\text{x}^3}+\Big({18}\times2{\text{x}}\Big)+\Big(18\times\frac{3}{\text{x}}\Big)$
$=\Big(8\text{x}^3+\frac{27}{\text{x}^3}+36\times\frac{54}{\text{x}}\Big)$
Hence The cube of $\Big(2\text{x}+\frac{3}{\text{x}}\Big)^3=\Big(8\text{x}^3+\frac{27}{\text{x}^3}+36\times\frac{54}{\text{x}}\Big)$
we know that $\text{a}=2\text{x},\text{b}=\frac{3}{\text{x}}$ By using $(a + b)^3$
formula $=8\text{x}^3+\frac{27}{\text{x}^3}+\frac{18\text{x}}{\text{x}}\Big(\frac{2}{\text{x}}+\frac{3}{\text{x}}\Big)$
$=8\text{x}^3+\frac{27}{\text{x}^3}+\frac{18\text{x}}{\text{x}}\Big(2{\text{x}}+\frac{3}{\text{x}}\Big)$
$=8\text{x}^3+\frac{27}{\text{x}^3}+\Big({18}\times2{\text{x}}\Big)+\Big(18\times\frac{3}{\text{x}}\Big)$
$=\Big(8\text{x}^3+\frac{27}{\text{x}^3}+36\times\frac{54}{\text{x}}\Big)$
Hence The cube of $\Big(2\text{x}+\frac{3}{\text{x}}\Big)^3=\Big(8\text{x}^3+\frac{27}{\text{x}^3}+36\times\frac{54}{\text{x}}\Big)$
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$ii.$ If $BL = BM$, prove that $AB = BC$.
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