Question

Plz solve this question.........fast ..... it's urgent........

Class 9 ....Ch-2......CCE questions.....R.S Agarwal......​

Answer 1

a²/bc+b²/ac+c²/ab=a³+b³+c³/abc—(1)

a³+b³+c³=(a+b+c)³−3a

(a+b)−3bc(b+c)−3ac(a+c)−6abc—(2)

Since,

a+b+c=0,b+c=−a,c+a=−b,a+b=−c—(3)

Substitute (3) in (2),

=>a³+b³+c³=03−3ab(−c)−3b(−a)−3ac(−b)−6abc

=>3abc+3abc+3abc−6abc=3abc—(4)

From

(1) and (4),a³+b³+c³/abc=3abc/abc=3

Ans: 3

Answer 2

Answer:

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