Question : If $a+b-c=20$ and $a^2+b^2+c^2=152$, find the value of $a^3+b^3-c^3+3abc$.
Option 1: 480
Option 2: 720
Option 3: 640
Option 4: 560
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Correct Answer: 560
Solution :
Given:
$a+b-c=20$ and $a^2+b^2+c^2=152$
$ a+b-c = 20$
Squaring both sides, we get,
⇒ $a^2 + b^2+c^2 + 2ab-2bc-2ca = 400$
⇒ $2(ab-bc-ca) = 400-152$
⇒ $ab-bc-ca = 124$
Now to find, $a^3+b^3-c^3+3abc$
$= (a+b-c)(a^2+b^2+c^2-(ab-bc-ca))$
$= 20\times (152-124)$
$= 20\times 28 = 560$
Hence, the correct answer is 560.
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