Find the set E of the values of x which the binomial expansion of $(3 - 4 x)^{\frac{3}{4}}$ is valid.

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Solution

${(3 - 4 x)}^{\frac{3}{4}}$
$= {(3)}^{\frac{3}{4}} {(1 - \frac{4 x}{3})}^{\frac{3}{4}}$
$= {(3)}^{\frac{3}{4}} {(1 + (- \frac{4 x}{3}))}^{\frac{3}{4}}$
Here $X = - \frac{4}{3} x$
As we know that, the expansion
${(1 + X)}^{n}$ ; When $n$ is a negative integer or a fraction is possible if
$- 1 < X < 1$
$- 1 < - \frac{4}{3} x < 1$
$- 1 \frac{4 x}{3} < - 1$
$- 3 < 4 x < 3$
$\frac{- 3}{4} < x < \frac{3}{4}$
$x \in (- \frac{3}{4}, \frac{3}{4})$
Hence for the given expansion to be valid set $E$ of values of $x$ is $(- \frac{3}{4}, \frac{3}{4})$ .

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