How do you solve $x(x^2-3x)^(1/3) + 2(x^2-3x)^(4/3) = 0$ ?

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Answer 1 · 2015-04-10T18:12:08

Temporarily replace $(x^2-3x)$ with $k$

$x(x^2-3x)^(1/3)+2(x^2-3x)^(4/3) = 0$

can be re-written (using the temporary substitution) as
$x*k^(1/3) + 2k*k^(1/3) = 0$

$(k^(1/3))*(x+ 2(x^2-3x)) = 0$

$(k^(1/3))*(2x^2 - 5x) = 0$

One of the terms
$k^(1/3) = root(3)(x^2-3x)$
or
$2x^2-5x$
must equal $0$

$root(3)(x^2-3x)$
can only equal $0$
if $x^2-3x = 0$
$rarr x = 0$
or
$x=3$

$2x^2-5x$
can only equal $0$
if $x=0$
or
$x = 5/2$

Therefore $x epsilon { 0, 2 1/2, 3}$

How do you solve x(x^2-3x)^(1/3) + 2(x^2-3x)^(4/3) = 0x(x^2-3x)^(1/3) + 2(x^2-3x)^(4/3) = 0?