Factor this equation

Question

Can someone factor this for me? $(x^{\frac{n}3}-a^{\frac{n}3})$

I am stuck on it. Let n be any natural number.

Answer 1

From your comment, it looks like you are being asked to find a "difference quotient". Subject the condition that the numbers being raised to fractional exponents need to be positive, you can treat this as

$$ \frac{x^{n/3} \ - \ a^{n/3}}{x \ - \ a } \ = \ \frac{{x^{n/3}} \ - \ a^{n/3}}{(x^{1/3})^3 \ - \ (a^{1/3})^3 } $$

$$ = \ \frac{{x^{n/3}} \ - \ a^{n/3}}{(x^{1/3} \ - \ a^{1/3}) \ \left[ (x^{1/3})^2 \ + \ (x^{1/3})(a^{1/3}) \ + \ (a^{1/3})^2 \right] } \ \ , $$

applying the "difference of two cubes".

Now the issue becomes what positive integer $ \ n \ $ is. If $ \ n = 1 \ , $ we'll just have

$$ \frac{1}{x^{2/3} \ + \ x^{1/3} \ a^{1/3} \ + \ a^{2/3} } \ \ , $$

after canceling numerator against denominator factor. For $ \ n = 2 \ , $ we'd have

$$ \frac{(x^{1/3} \ - \ a^{1/3}) \ ( x^{1/3} \ + \ a^{1/3})}{(x^{1/3} \ - \ a^{1/3}) \ \left[ (x^{1/3})^2 \ + \ (x^{1/3})(a^{1/3}) \ + \ (a^{1/3})^2 \right] } \ \ , $$

$$ = \ \frac{ x^{1/3} \ + \ a^{1/3}}{ x^{2/3} \ + \ x^{1/3} \ a^{1/3} \ + \ a^{2/3} } \ \ , $$

using "difference of two squares" in the numerator first. You should always be able to cancel the "difference of cube roots" factors between the numerator and denominator, though the expression may cease to be so tidy for large values of $ \ n \ . $