Question : If $a^2+b^2+c^2=160$ and $a+b+c=16$, find $a b+b c+c a$.
Option 1: 96
Option 2: 84
Option 3: 48
Option 4: 42
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Correct Answer: 48
Solution : Given, $a^2+b^2+c^2=160$ and $a+b+c=16$ We know, $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ ⇒ $16^2 = 160 + 2(a b+b c+c a)$ ⇒ $2(a b+b c+c a)=256-160$ ⇒ $ab+bc+ca=48$ Hence, the correct answer is 48.
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Question : If $(a+b+c)=17$ and $\left(a^2+b^2+c^2\right)=101$, then find the value of $(a-b)^2+(b-c)^2+(c-a)^2$.
Option 1: 12
Option 2: 14
Option 3: 10
Option 4: 16
Question : If $9 a+9 b+9 c=81$ and $4 a b+4 b c+4 c a=160$, then what is the value of $6 a^2+6 b^2+6 c^2$?
Option 1: 1
Option 2: 3
Option 3: 4
Option 4: 6
Question : In a triangle ABC; 8$\angle$A = 6$\angle$B = 3$\angle$C. What are the degree measures of $\angle$ A, $\angle$ B, and $\angle$C?
Option 1: $48^{\circ}, 96^{\circ}, \text{and } 36^{\circ}$
Option 2: $36^{\circ}, 96^{\circ}, \text{and } 48^{\circ}$
Option 3: $36^{\circ}, 48^{\circ}, \text{and } 96^{\circ}$
Option 4: $96^{\circ}, 48^{\circ}, \text{and } 36^{\circ}$
Question : Simplify: $(2 a-3 b-c)^2-(a+2 b+c)^2$
Option 1: $3 a^2+5 b^2-16 a b+2 b c-8 a c$
Option 2: $3 a^2+5 b^2-8 a b+2 b c-6 a c$
Option 3: $3 a^2+5 b^2-16 a b-6 a c$
Option 4: $3 a^2+5 b^2-16 a b+2 b c-6 a c$
Question : If $a+b+c=5$ and $a^2+b^2+c^2=15$, then find the value of $a^3+b^3+c^3-3 a b c-27$.
Option 1: 23
Option 2: 27
Option 3: 25
Option 4: 21
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