Please answer this question.

Please answer this question. 33. If ax + by + c = O, bx + cy + a = O and cx + ay + b = 0 passes through the same ordered pair, then show that a3 + b3 + c3 — 3abc = 0.

Hello,

Thanks for A2A

This is quite simple:

Let

ax+by+c=0 ?..(1)

bx+cy+a=0 ?..(2)

cx+ay+b=0 ?..(3)

Now add all three equation

ax+by+c+bx+cy+a+cx+ay+b=0

ax+ay+a+bx+by+b+cx+cy+c=0

a(x+y+1)+b(x+y+1)+c(x+y+1)=0

Now put common term outside

(x+y+1)[a+b+c]=0

We that rule if?A*B=0?then either A=0 or B=0 so if

=> a+b+c=0

=> a+b= -c ??(4)

Now calculate cube both side

=> ( a+b)?=(-c)?

=> a?+b?+3ab(a+b)= -c?

Now from equation (4) a+b=-c

=> a?+b?-3abc= -c?

=> a?+b?+c? = 3abc (prove)

Thanks!