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
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.
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!!!