Solving equations with fractional indices How would I go about solving an equation like this?
$3^{4/3}b^{5/3} - b^3 = 1$
I thought about rearranging to get $3^{4/5}b = (1 + b^3)^{3/5}$, but that didn't seem to lead anywhere as I couldn't multiply out the term on the RHS.
 A: Hint
Consider the function $$f(b)=3 \sqrt[3]{3} b^{5/3}-b^3-1$$ and its derivatives  $$f'(b)=5 \sqrt[3]{3} b^{2/3}-3 b^2$$  $$f''(b)=\frac{10}{3^{2/3} \sqrt[3]{b}}-6 b$$ You can notice that the first derivative cancels for $b_1=0$ and $b_2=\frac{5^{3/4}}{\sqrt{3}}$. You can also notice that $f(0)=-1$ and $$f(b_2)=\frac{20 \sqrt[4]{5}}{3 \sqrt{3}}-1\approx 4.7556$$ is positive; so you know that there is one root between $b_1$ and $b_2$. At $b_2$, the second derivative is $$f''(b_2)=-\frac{4\ 5^{3/4}}{\sqrt{3}}$$ is negative so $b_2$ corresponds to a maximum of the function. Now, for large values of $b$, the function is driven by $-b^3$ then it will go to $-\infty$; so there is a second root after $b_2$. ou can now notice that $f(3)=-1$; so the the second root very close to $3$. Finding this root accurately requires numerical methods such as Newton (since I do not know if you already heard about, I shall not address this aspect - if you want, let me know and I shall elaborate). Concerning the first root, by inspection, you could notice that it is between $\frac{1}{3}$ and  $\frac{1}{2}$.
A: Set $b=x^3$ and you get an ordinary algebraic equation that you can solve by standard numerical methods.
$$3^{4/3}x^5 - x^9 = 1.$$
There are three real roots in $x$, close to $-1.4$, $0.75$ and $1.4$.
