Math, asked by bandhubajaj5838, 1 year ago

If the lines ax+by+c=0, bx+cy+a=0 and cx+ay+b=0 be concurrent then prove that a3+b3+c3-3abc=0

Answers

Answered by Varshu123
86

ax + by + c = 0


bx + cy + a = 0


cx + ay + b = 0


Adding all the above equations, we get,


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


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


Assuming, (a + b + c) = 0, we get,


a + b = -c ... (1)


Cubing both sides,


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


a^3 + b^3 + 3ab(-c) = -c^3 ... from eq. (1)


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


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


If, (x + y + 1), then a,b and c doesn't satisfy the given equation.

Answered by shriya1707
9

Hey there,

ax + by + c = 0

bx + cy + a = 0

cx + ay + b = 0

Adding all the above equations, we get,

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

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

Assuming, (a + b + c) = 0, we get,

a + b = -c ... (1)

Cubing both sides,

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

a^3 + b^3 + 3ab(-c) = -c^3 ... from eq. (1)

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

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

If, (x + y + 1), then a,b and c doesn't satisfy the given equation.

Hope it helps you!!!

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