Question

if a+b+c=0 show that a²/bc+b²/ac+c²/ab
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Answer 1

Question

If (a+b+c) = 0 , then show that a²/bc + b²/ca + c²/ab = 3 .

Solution

Given:-

• a + b + c = 0 ------------(1)

Show that :-

• a²/bc + b²/ca + c²/ab = 3

Explanation

We know,

★ (a+b+c)³ = a³+b³+c³+3(a+b)(b+c)(c+a)

keep value by equation (1),

➠ 0³ = a³+b³+c³+3(a+b)(b+c)(c+a)

➠ a³+b³+c³ = -3(a+b)(b+c)(c+a)

By, equ(1),

➠ c = -(a+b) --------(2)

keep these value in above ,

➠ a³+b³+c³ = -3(a+b).(b-a-b).(-a-b+a)

➠ a³+b³+c³ = -3.(a+b).(-a).(-b)

Again,

Again,By, equ(1),

➠ a+b = -c ---------(3)

So,

➠ a³+b³+c³ = -3.(-c).(-a).(-b)

➠ a³+b³+c³ = 3abc

Divided by "abc" in both sides.

➠ (a³+b³+c³)/abc = 3abc/abc

➠ a³/abc + b³/abc + c³/abc = 3

➠ a²/bc + b²/ca + c²/ab = 3

That's proved.

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