The solution of $x^{1 / 3} + (2 x - 3)^{1 / 3} = [3 [x - 1]]^{1 / 3}$ is

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\frac{3}{2}

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Solution

The correct options are
A 0
B $\frac{3}{2}$
C 1
The solution of $a^{1 / 3} + b^{1 / 3} = (a + b)^{1 / 3}$ is true,

when either $a = 0$ or $b = 0 o r a + b = 0$
Here,
$x^{1 / 3} + (2 x - 3)^{1 / 3} = [3 [x - 1]]^{1 / 3}$

$x^{1 / 3} + (2 x - 3)^{1 / 3} = [3 x - 3]]^{1 / 3}$

$x = 0, 2 x - 3 = 0, 3 x - 3 = 0$

$\Rightarrow x = 0, \frac{3}{2}, 1.$

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