Question : If $(a+b+c)=20$ and $a^2+b^2+c^2=152$, find the value of $a^3+b^3+c^3-3 abc$.
Option 1: 560
Option 2: 640
Option 3: 480
Option 4: 720
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Correct Answer: 560
Solution :
Given: $(a+b+c)=20$ and $a^2+b^2+c^2=152$
We know the identity,
$(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$
$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
⇒ $20^2=152+2(ab + bc + ca)$
⇒ $ab + bc + ca=\frac{400-152}{2}$
⇒ $ab + bc + ca= 124$ .........(i)
Also,
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
From (i),
$a^3+b^3+c^3-3abc=(20)(152-124)$
⇒ $a^3+b^3+c^3-3abc=560$
Hence, the correct answer is 560.
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