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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