Question

If a,b,c are real numbers such that a+b+c= 0 & a²+ b²+ c²= 1, find (3a+5b-8c)²+(-8a+3b+5c)²+(5a-8b+3b)². Steps plz.​

Answer 1

Answer:

147

Step-by-step explanation:

3a+5b−8c)2=9a2+25b2+64c2+30ab−80bc−48ac

(−8a+3b+5c)2=64a2+9b2+25c2−48ab+30bc+−80ac

(5a−8b+3c)2=25a2+64b2+9c2−80ab−48bc+30ac

adding all three

=9+25+64+30(ab+bc+ca)−80(ab+bc+ca)−48(ab+bc+ca)

=98−98(ab+bc+ca)

=98−98(((a+b+c)2−a2−b2−c2)/2)

=98+49

=147