Math, asked by mopidevimanideep, 5 months ago

If a+b+c80, ax +by+c= 0 bx + cy+ a = 0, cx + ay+b=0
a² +6² +6²
are concurrent then ab + bc+ca​

Answers

Answered by ifratgul333
0

Answer:

ax+by+c=0

bx+cy+a=0

cx+ay+b=0

Adding all the above equations, we get,

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

ax+ay+a+bx+by+b+cx+xy+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

.........fromeq.(1)

a

3

+b

3

−3abc=−c

3

a

3

+b

3

+c

3

=3abc

Answered by sypraveen141004
2

Answer:

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