I’m not sure how to solve 3d. College calculus 1
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Step 1
Given;
[tex]f(x)=(x^2+5)^3[/tex]Required; To simplify
[tex]\frac{f(x)-f(0)}{x},\text{ x}\ne0[/tex]Step 2
[tex]\frac{(x^2+5)^3-(0^2+5)^3}{x}[/tex][tex]\mleft(a+b\mright)^3=a^3+3a^2b+3ab^2+b^3---(apply\text{ p}\operatorname{erf}ect\text{ cube formula)}[/tex][tex](x^2+5)^3=(x^2)^3+3(x^2)^2(5)+3x^2(5^2)+5^3[/tex][tex](x^2+5)^3=x^6+15x^4+75x^2+125[/tex][tex]\frac{(x^6+15x^4+75x^2+125)-125}{x}[/tex][tex]\begin{gathered} \frac{x^6+15x^4+75x^2+125-125}{x} \\ \frac{x^6+15x^4+75x^2}{x} \\ \frac{f(x)-f(0)}{x}=x^5+15x^3+75x \end{gathered}[/tex]Hence if we factorize we get;
[tex]x(x^4+15x^2+75)[/tex]Therefore;
[tex]\frac{f(x)-f(0)}{x}=x(x^4+15x^2+75)[/tex]