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# help with algebra

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Let $a$ and $b$ be real numbers such that $a^3 + 3ab^2 = 679$ and $a^3 - 3ab^2 = 615.$ Find $a - b.$
Thanks!

Jun 13, 2023

#1
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To find the value of $a - b$, we can subtract the second equation from the first equation:

\begin{align*}
(a^3 + 3ab^2) - (a^3 - 3ab^2) &= 679 - 615 \
6ab^2 &= 64 \
ab^2 &= \frac{64}{6} \
ab^2 &= \frac{32}{3}
\end{align*}

Now, we can substitute this value of $ab^2$ into one of the original equations. Let's use the first equation:

\begin{align*}
a^3 + 3ab^2 &= 679 \
a^3 + 3\left(\frac{32}{3}\right) &= 679 \
a^3 + 32 &= 679 \
a^3 &= 679 - 32 \
a^3 &= 647
\end{align*}

Now, we can take the cube root of both sides to solve for $a$:

\begin{align*}
a &= \sqrt{647} \
a &= 7
\end{align*}

Finally, we can substitute the value of $a$ back into one of the original equations to find the value of $b$. Let's use the first equation again:

\begin{align*}
7^3 + 3(7)b^2 &= 679 \
343 + 21b^2 &= 679 \
21b^2 &= 679 - 343 \
21b^2 &= 336 \
b^2 &= \frac{336}{21} \
b^2 &= 16 \
b &= \pm 4
\end{align*}

Since we are interested in finding $a - b$, we take the positive value of $b$:

\begin{align*}
a - b &= 7 - 4 \
a - b &= 3
\end{align*}

Therefore, $a - b = \boxed{3}$.   MyBKExperience Survey Code

Jun 13, 2023