if a+b=4 and a^3+b^3=28 ,find a*b

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on Nov 23, 2024 at 9:03 am



You asked:

Given the following equations: \[\begin{aligned}a + b &= 4\\ {a}^{3} + {b}^{3} &= 28\end{aligned}\] Evaluate \(a b\).

MathBot Answer:

The system of equations has \(2\) solutions.

\[a = 1, b = 3\]\[a = 3, b = 1\]

By substituting these values into the expression, we find that: $$\begin{aligned}a b &= 3\end{aligned}$$

Solve \(a + b = 4\) for \(a\). \[a = 4 - b\]Substitute \(4 - b\) for \(a\) in \(a^{3} + b^{3} = 28\) and simplify. $$\begin{aligned}a^{3} + b^{3} &amp= 28 \\ \left(4 - b\right)^{3} + b^{3} &= 28 \\ b^{2} - 4 b &= -3 \\b^{2} - 4 b + 3 &= 0 \\ \left(b - 3\right) \left(b - 1\right) &= 0 \\ b = 1&, b = 3\end{aligned}$$Substitute \(1\) into \(a + b = 4\) to solve for \(a\). \[\begin{aligned}a + 1 &= 4\\a &= 3\end{aligned}\]This yields the following solution. $$\begin{aligned}a = 3,\,b = 1\end{aligned}$$Substitute \(3\) into \(a + b = 4\) to solve for \(a\). \[\begin{aligned}a + 3 &= 4\\a &= 1\end{aligned}\]This yields the following solution. $$\begin{aligned}a = 1,\,b = 3\end{aligned}$$

By substituting these values into the expression, we find that: $$\begin{aligned}a b &= 3\end{aligned}$$

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