Question

A-b=3.b-c=5.find a2+b2+c2-ab-bc-ca?

Answer 1

hope this helps you.....

Answer 2

Answer:

The value of a2+b2+c2-ab-bc-ca is 49.

Step-by-step explanation:

Given information:

$a-b=3$             .... (1)

$b-c=5$             .... (2)

Add (1) and (2).

$a-b+b-c=3+5$

$a-c=8$              .... (3)

$a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}[2a^2+2b^2+2c^2-2ab-2bc-2ca]$

$a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}[(a^2-2ab-b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)]$

$a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}[(a-b)^2+(b-c)^2+(a-c)^2]$

Using (1), (2) and (3), we get

$a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}[(3)^2+(5)^2+(8)^2]$

$a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}[98]=49$

Therefore the value of a2+b2+c2-ab-bc-ca is 49.