Question
Write ${\left( {\frac{3}{2}x + 1} \right)^3}$ in expanded form.
✓
Answer
${\left( {\frac{3}{2}x + 1} \right)^3}$
We know that ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$
${\left( {\frac{3}{2}x + 1} \right)^3} = {\left( {\frac{3}{2}x} \right)^3} + {\left( 1 \right)^3} + 3 \times \frac{3}{2}x \times 1\left( {\frac{3}{2}x + 1} \right)$
$= \frac{{27}}{8}{x^3} + 1 + \frac{9}{2}x\left( {\frac{3}{2}x + 1} \right)$$ = \frac{{27}}{8}{x^3} + \frac{{27}}{4}{x^2} + \frac{9}{2}x + 1.$
Therefore, the expansion of the expression ${\left( {\frac{3}{2}x + 1} \right)^3}$ is $\frac{{27}}{8}{x^3} + \frac{{27}}{4}{x^2} + \frac{9}{2}x + 1$
We know that ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$
${\left( {\frac{3}{2}x + 1} \right)^3} = {\left( {\frac{3}{2}x} \right)^3} + {\left( 1 \right)^3} + 3 \times \frac{3}{2}x \times 1\left( {\frac{3}{2}x + 1} \right)$
$= \frac{{27}}{8}{x^3} + 1 + \frac{9}{2}x\left( {\frac{3}{2}x + 1} \right)$$ = \frac{{27}}{8}{x^3} + \frac{{27}}{4}{x^2} + \frac{9}{2}x + 1.$
Therefore, the expansion of the expression ${\left( {\frac{3}{2}x + 1} \right)^3}$ is $\frac{{27}}{8}{x^3} + \frac{{27}}{4}{x^2} + \frac{9}{2}x + 1$
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