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how do you express (a + b) to the exponent 0.25 in terms of a and b separately? For instance (2 + 4) to the 0.25 = 1.5651 but how would you solve

(2 + unkown) to the 0.25 = 1.5651?

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    Note that $(2+4)^{0.25}=\sqrt[4]6 \neq 1.5651$. It is close, but not exact.2012-12-03

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You can't in the general cases. The simplest way it can be written is $\sqrt[4]{a+b}$ or $(a+b)^{\frac{1}{4}}$.

To solve an equation like $(2+x)^{\frac{1}{4}}=1.5651,$ raise both sides of the equation to the $4$th power. You arrive at $x=1.5651^{4}-2.$

The idea in all cases such as this is removal of the radical by raising both sides of the equation to the power adequate to remove the radical. In general, $x^{\frac{1}{n}}=y$ would require us to raise both sides to the $n$th power to solve the equation for $x$.

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    Yeah, my bad. @JuliánAguirre2012-12-03
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This is Isaac Newon's (generalized) binomial theorem.

\begin{align} (a+b)^{0.25} & = a^{0.25} + 0.25a^{0.25-1}b + \frac{(0.25)(0.25-1)}{2} a^{0.25-2}b^2 + {} \\[12pt] & {}+ \frac{(0.25)(0.25-1)(0.25-2)}{6} a^{0.25-3} b^3+ \cdots\cdots \end{align} This works if $|a|>|b|$. If the inequality is reversed, then interchange the roles of $a$ and $b$.

(The denominators are the factorials $1,1,2,6,24,120,\ldots$ .