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# Algebra

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Let a and b be complex numbers. If a + b = 4 and a^2 + b^2 = 6 + ab, then what is a^3 + b^3?

Jul 1, 2024

#1
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We don't actually have to find what a and b is.

First, let's note that we have \((a+b)^2 = a^2+2ab+b^2\)

Squaring both sides of the first equation, we find that

\(a^2+2ab+b^2 = 16\)

Subtracting the second equation from this equation we just computed, we have

\(3ab = 10\\ ab = 10/3\)

Now, let's note something else real quick.

We have that \((a+b)^3=a^{3}+3a^{2}b+3ab^{2}+b^{3} = a^3 + 3ab(a+b)+b^3\)

Since we also know that \((a+b)^3 = 4^3=64\), we have the equation \( a^3 + 3ab(a+b)+b^3=64\)

Wait! we already have all the numbers needed to isolate a^3 and b^3. We just found what ab was and a+b is given, so we have

\(a^3+b^3=64-3(10/3)(4) \\a^3+b^3 = 64-40\\ a^3+b^3=24\)

So 24 is our final answer.

Thanks! :)

Jul 1, 2024
#2
+302
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For this question, use formula: (a+b)^2 = a^2+b^2+2ab

since a+b=4

a^2+b^2+2ab = 16

a^2+b^2 = 6 + ab

substitute:

6 + ab + 2ab = 16

3ab + 6 =16

3ab = 10

ab= 10/3

a^3 + b^3 = (a+b)(a^2-ab+b^2)

a+b = 4, a^2+b^2 = 6 + 10/3 = 28/3 , and -ab = -10/3

28/3 - 10/3 = 6

=> 4*6 = 24

here's another way to understand :)

(good job, @NotThatSmart)

edited by aboslutelydestroying  Jul 1, 2024
#3
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Very smart way to do it!

I totally forgot about factoring a^3 + b^3 as \((a+b)(a^2-ab+b^2)\)

It makes a lot of sense and is probably the most efficient way to complete this problem.

Subsituting in \((a+b)^3=a^{3}+3a^{2}b+3ab^{2}+b^{3} = a^3 + 3ab(a+b)+b^3\) work, I geuss, but good work!

Both solutions do eventually reach the desired goal of 24, so that's awesome. :)

Good job to you too, @aboslutelydestroying!

NotThatSmart  Jul 1, 2024
edited by NotThatSmart  Jul 1, 2024
edited by NotThatSmart  Jul 1, 2024