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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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
Question : If $a+b+c=5, a^3+b^3+c^3=85$, and $abc=25$, then find the value of $a^2+b^2+c^2-a b- bc - ca$.
Option 1: 2
Option 2: 4
Option 3: 6
Option 4: 8
Question : If $(a^3+b^3+c^3-3 a b c)=405$ and $(a-b)^2+(b-c)^2+(c-a)^2=54$, find the value of $(a + b + c)$.
Option 1: 15
Option 2: 45
Option 3: 9
Option 4: 27
Question : If $a, b, c$ are real numbers and $a^{2}+b^{2}+c^{2}=2(a-b-c)-3,$ then the value of $a+b+c$ is:
Option 1: –1
Option 2: 1
Option 3: 3
Option 4: 0
Question : If $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}$ where $a \neq b\neq c\neq 0$, then the value of $a^{2}b^{2}c^{2}$ is:
Option 2: abc
Option 3: 0
Option 4: 1
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