Factoring under a 4th power

Question

Show that $$\left( a + b \right) ^ 4 = b^4 \left( \frac{a}{b} + 1 \right) ^4$$

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Its clear without an exponent

$$\left( a + b \right) = b \left( \frac{a}{b} + 1 \right) $$

but I'm not sure why should it hold to the $n$th power

Answer 1

Hints:

(1) If $a = b$, then $a^n = b^n$ for all nonnegative integers $n$.

(2) $(ab)^n = {a^n}{b^n}$

Can you take it from here?

Answer 2

$\left( a + b \right) ^ 4 = \left(b \left( \frac{a}{b} + 1 \right)\right)^4 =b^4 \left( \frac{a}{b} + 1 \right) ^4$ assuming comutative multiplication