If $a≠b$, and $a²=5a-7$ and $b²=5b-7$, then $a³+b³=?$ How can I find the answer? I found out that $a$ and $b$ are not real numbers. Can anyone please teach me? thank you!!
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$a$ and $b$ are both solutions of $x^2-5a+7=0$, which tells you their sum and product, and also tells you $a^2+b^2$. And $a^3+b^3$ can be factored.... – Gerry Myerson Nov 20 '16 at 03:28
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You can do it in this way:
$a^3+b^3=(a+b)^3-3ab(a+b)$
Easily you can know that $a,b$ are two solutions of equation $x^2-5x+7=0$, so you can know that $a+b=5$ and $ab=7$, so you can get
$a^3+b^3=5^3-3\times7\times5=20$
duanduan
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An alternative solution, without Vieta's formulas or binomial expansion or sum of cubes factors.
First subtract the two equations to get $a^2-b^2 = 5(a-b)$. Since it's given that $a \ne b\,$, divide by $a-b$ which leaves $a+b=5$.
Next note that $a^2=5a-7\,$, so $\,a^3= 5a^2-7a=5(5a-7)-7a=18a -35\,$.
Same works for $b$, then adding the two: $\,a^3+b^3=18(a+b)-70=18 \cdot 5 - 70 = 20$.
dxiv
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