Question

If
$b = {a}^{ \frac{1}{3} } - {a}^{ \frac { - 1}{3} }$
then show that
$b ^{3} + 3b = a - \frac{1}{a}$
​

Answer 1

Step-by-step explanation:

If

$b = {a}^{ \frac{1}{3} } - {a}^{ \frac { - 1}{3} }$

then show that

$b ^{3} + 3b = a - \frac{1}{a}$

If

$b = {a}^{ \frac{1}{3} } - {a}^{ \frac { - 1}{3} }$

then show that

$b ^{3} + 3b = a - \frac{1}{a}$

If

$b = {a}^{ \frac{1}{3} } - {a}^{ \frac { - 1}{3} }$

then show that

$b ^{3} + 3b = a - \frac{1}{a}$

If

$b = {a}^{ \frac{1}{3} } - {a}^{ \frac { - 1}{3} }$

then show that

$b ^{3} + 3b = a - \frac{1}{a}$

ex]b ^{3} + 3b = a - \frac{1}{a} [/tex]