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Solve (x+4)^3 + (x+5)^3 = (x+7)^3 + (x-2)^3

 Jun 21, 2021
 #1
avatar+287 
+2

I will avoid expanding any of the cubed expression.  Define a polynomial $p$ to be $p(x)=(x+4)^3 + (x+5)^3 - (x+7)^3 - (x-2)^3$.  Note that $p$ is at most quadratic.  Thus, $p(x)=ax^2+bx+c$ for some real $a,b,c$.  (In fact, $a,b,c$ are integers, but it's enough to say they are real.) We have
\begin{eqnarray*}
    p(0) = c &=& 4^3+5^3-7^3 - (-2)^3 = -146 \\
    p(1) = a+b+c &=& 5^3+6^3-8^3-(-1)^3 = -170 \\
    p(-1) = a-b+c &=& 3^3+4^3-6^3-(-3)^3 = -98
\end{eqnarray*}
Solving for $a$, $b$, and $c$, we have $a=12$, $b=-36$, and $c=-146$, giving $p(x)= 12x^2-36x-146$.   So
\[
    x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} = \dfrac{3}{2}\pm\dfrac{1}{6\sqrt{519}}.
\]


 

 Jun 22, 2021
 #2
avatar+502 
+1

idk if this is right

 

\(x=\frac{9+\sqrt{519}}{6},\:x=\frac{9-\sqrt{519}}{6}\)

 Jun 22, 2021

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