Question

Question 3 (2 marks)
Q3. If ab+bc+ca=71 and a+b+c=15, then the value of a2+b2+c2 is equal to:​

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• ab + bc + ac = 71
• a + b + c = 15

${\sf{\green{\underline{\large{To\:find}}}}}$

• Then
• a²+b²+c²

${\sf{\pink{\underline{\Large{Explanation}}}}}$

Identity

=>(a+b+c)²=a²+b²+c²+2(ab + bc + ac)

¶utting the values

=>(a+b+c)²=a²+b²+c²+2(ab + bc + ac)

=>(15)²=a²+b²+c²+2(71)

=>225 =a²+b²+c²+142

=>225-142=a²+b²+c²

=>a²+b²+c²= 83

Additional Information

$\boxed{\boxed{\sf\red{(a+b)^2=a^2+b^2+2ab}}}$

$\implies\tt{X^3+\dfrac{1}{X^3)}=(X+\dfrac{1}{x})(X^2+\dfrac{1}{X^2}-\frac{1}{X}\times{X)}}$

Answer 2

Answer:

Step-by-step explanation:

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