Question

If a+b+c= 0 prove that a(a2 - bc) + b(b2 - ca) + c(c2- ab) = 0.

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Answer 1

Step-by-step explanation:

a(a2 - bc) + b(b2 - ca) + c(c2- ab)

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

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

=(a+b+c)×(a^2+b^2+c^2-ab-bc-ca)

=0 [since a+b+c = 0]

Answer 2

a(a² - bc) + b(b²- ca) + c(c² - ab) = 0

GIVEN

a + b + c = 0

TO PROVE

a(a² - bc) + b(b²- ca) + c(c² - ab) = 0

SOLUTION

We can simply solve the above problem as follows;

We have to prove,

a(a² - bc) + b(b²- ca) + c(c² - ab) = 0

Let us simplify the given expression,

a³ - abc + b³ - abc + c³ - abc

Adding the like terms;

a³ + b³ + c³ - 3abc

We know that;

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

Putting (a+b+c) as 0 in the above equation;

= 0 × (a² + b² + c² - ab - bc - ca)

Since anything multiplied to zero is zero;

= 0

Therefore,

a(a² - bc) + b(b²- ca) + c(c² - ab) = 0

a(a² - bc) + b(b²- ca) + c(c² - ab) = 0 Hence, Proved.

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