How do you simplify $ (x^{1/3} + x^{-1/3})^2 $?
Answer 1
Curtis Vasquez
To simplify
$ (x^{1/3} + x^{-1/3})^2 $
expand using the binomial theorem:
$
(a + b)^2 = a^2 + 2ab + b^2
$
Here:
- $ a = x^{1/3} $
- $ b = x^{-1/3} $
So:
$
(x^{1/3} + x^{-1/3})^2 = (x^{1/3})^2 + 2(x^{1/3})(x^{-1/3}) + (x^{-1/3})^2
$
Simplify each term:
- $ (x^{1/3})^2 = x^{2/3} $
- $ 2(x^{1/3})(x^{-1/3}) = 2(x^{(1/3 – 1/3)}) = 2(x^0) = 2 $
- $ (x^{-1/3})^2 = x^{-2/3} $
Combine the results:
$
x^{2/3} + 2 + x^{-2/3}
$
Thus, the simplified expression is:
$
x^{2/3} + 2 + x^{-2/3}
$
Answer 2
Emma Smith
Expand $ (x^{1/3} + x^{-1/3})^2 $ using the binomial theorem:
$$
a^2 + 2ab + b^2
$$
Here $ a = x^{1/3} $ and $ b = x^{-1/3} $. Simplify:
- $ (x^{1/3})^2 = x^{2/3} $
- $ 2(x^{1/3})(x^{-1/3}) = 2 $
- $ (x^{-1/3})^2 = x^{-2/3} $
Final result:
$
x^{2/3} + 2 + x^{-2/3}
$
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