The value of x such that f(x) = g(x) is x = 3
<h3>Quadratic equation</h3>
Given the following expressions as shown
f(x) = x^3-3x^2+2 and;
g(x) = x^2 -6x+11
Equate the expressions
x^3-3x^2+2 = x^2 -6x+11
Equate to zero
x^3-3x^2-x^2+2-11 = 0
x^3-3x^2-x^2 + 6x - 9 = 0
x^3-4x^2+6x-9 = 0
Factorize
On factorizing the value of x = 3
Hence the value of x such that f(x) = g(x) is x = 3
Learn more on polynomial here: brainly.com/question/2833285
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I wonder if you mean to write ![\sqrt[3]{256}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B256%7D)
in place of 
...
If you meant what you wrote, then we have
If you meant to write (the cube root of 256), then we could go on to have
![\sqrt[3]{256}=\sqrt[3]{16^2}=\sqrt[3]{(4^2)^2}=\sqrt[3]{4^4}=\sqrt[3]{4^3\cdot4}=4\sqrt[3]4](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B256%7D%3D%5Csqrt%5B3%5D%7B16%5E2%7D%3D%5Csqrt%5B3%5D%7B%284%5E2%29%5E2%7D%3D%5Csqrt%5B3%5D%7B4%5E4%7D%3D%5Csqrt%5B3%5D%7B4%5E3%5Ccdot4%7D%3D4%5Csqrt%5B3%5D4)
