Question

If a2+b2+c2= 74 and ab+bc+ca=61 find (a+b+c)2

Answer 1

The value of $(a+b+c)^2=196$

Step-by-step explanation:

Given by question,

$a^2+b^2+c^2=74$ and $ab+bc+ca=61$      .....(1)

To find, $(a+b+c)^2=?$

We know that,

$(x+y+z)^2=x^{2}+y^{2} +z^{2} +2(xy+yz+zx)$

∴ $(a+b+c)^2=a^{2}+b^{2} +c^{2} +2(ab+bc+ca)$

Using (1), we get

$(a+b+c)^2=74 +2(61)$

⇒$(a+b+c)^2=74 +122=196$

∴ $(a+b+c)^2=196$

Hence, the value of $(a+b+c)^2=196$

Answer 2

Answer:

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