# Powers of sums/Fractional exponents

 P: 20 for your first problem, as it stands (x + 5)3 = 3x + 15 but i think you meant to say $$(x+5)^3$$ you can do this by applying the following twice $$(a+b)^2=a^2+2ab+b^2$$ $$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$ so $$(x+5)^3=x^3+15x^2+75x+125$$
 P: 1,157 You first problem is a bit strange as it just yields solutions at x=-5 Your second question can be broken down a bit, basically by employing $$x^{ab}=(x^a)^b$$ and $$x^{-a}=1/x^a$$ i.e. $$(-27)^{-2/3}=(-1)^{-2/3}(27)^{-2/3}=(-1)^{-2/3}((27)^{-2})^{1/3}=(-1)^{-2/3}((1/27)^2)^{1/3}=(-1)^{-2/3}(1/729)^{1/3}=(-1)^{-2/3}(1/9)$$ The $$(-1)^{-2/3}$$ bit requires you to know a bit about complex numbers too...
 Sci Advisor HW Helper PF Gold P: 12,016 Nimz: And, what, may I ask, is $$(-1)^{-\frac{4}{6}}$$ equal to? Do you still think that no knowledge of complex numbers is needed?
 P: 1,157 Exactly, last time I looked... $$-1=e^{i\pi}$$ ie. $$(-1)^{-2/3}=e^{-2i\pi/3}=\cos(2\pi/3)-i\sin(2\pi/3)\approx-0.5-i0.866$$
 Sci Advisor HW Helper P: 11,896 Actually $$-1=e^{i\left(2k+1\right)\pi} , \forall{k\in\mathbb{Z}}$$ Daniel.