Question

find the point where line 3bx+ay=(a^(3)+b^(3)+c^(3))/(bc) cuts the y- axis.given that a+b+c=0

Answer 1

Answer:

Given that,

Equation :y=2x+3

The point at which it cuts y-axis has x-coordinate as 0

∴y=2(0)+3=3

Hence, it cuts y-axis at (0,3)

Step-by-step explanation:

See,

If a + b + c = 0

Then,

a + b + c = 0

a + b = - c -- (I)

(a + b)^3 = (- c)^3

a^3 + b^3 + 3ab (a + b) = - c^3

from equation (I) a + b = - c

a^3 + b^3 + 3ab (- c) = - c^3

a^3 + b^3 - 3abc = - c^3

a^3 + b^3 + c^3 = 3abc

Hence,

a^3 + b^3 + c^3 ‘not equal’ 0

but,

a^3 + b^3 + c^3 ‘equal’ 3abc

Answer 2

Step-by-step explanation:

hope it helps you.your answer is in the attachment