Question 824097
f(x)= x^3-x+1


f(h)= h^3-h+1 ... replace EVERY copy of x with h


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f(x)= x^3-x+1


f(-h)= (-h)^3-(-h)+1 ... replace EVERY copy of x with -h


f(-h)= -h^3-(-h)+1


f(-h)= -h^3+h+1


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f(h)-f(-h) = (h^3-h+1) - (-h^3+h+1)


f(h)-f(-h) = h^3-h+1 + h^3 - h - 1


f(h)-f(-h) = 2h^3-2h


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So...


(f(h)-f(-h))/2h


turns into


(2h^3-2h)/2h


then you factor out 2h in the numerator to get


(2h(h^2-1))/2h


and then cancel out the '2h' terms to get


h^2 - 1


and then you can optionally factor to get


(h-1)(h+1)


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So the answer is either <font color="red">h^2 - 1</font> or <font color="red">(h-1)(h+1)</font>


They are equivalent. The second expression is the factored form of the first (using the difference of squares rule).