Question

verify that,27a cube +b cube +c cube =(3a+b+c){9a square +b square +c square -3ab-bc-3ac}

Answer 1

Taking RHS of the identity: 
(a + b + c)(a2 + b2 + c2 - ab - bc - ca ) 

Multiply each term of first polynomial with every term of second polynomial, as shown below: 
= a(a2 + b2 + c2 - ab - bc - ca ) + b(a2 + b2 + c2 - ab - bc - ca ) + c(a2 + b2 + c2 - ab - bc - ca ) 

= { (a X a2) + (a X b2) + (a X c2) - (a X ab) - (a X bc) - (a X ca) } + {(b X a2) + (b X b2) + (b X c2) - (b X ab) - (b X bc) - (b X ca)} + {(c X a2) + (c X b2) + (c X c2) - (c X ab) - (c X bc) - (c X ca)} 

Solve multiplication in curly braces and we get: 
= a3 + ab2 + ac2 - a2b - abc - a2c + a2b + b3 + bc2- ab2 - b2c - abc + a2c + b2c + c3 - abc - bc2 - ac2 

Rearrange the terms and we get: 
= a3 + b3 + c3 + a2b - a2b + ac2- ac2 + ab2 - ab2 + bc2 - bc2 + a2c - a2c + b2c - b2c - abc - abc - abc 

Above highlighted like terms will be subtracted and we get: 
= a3 + b3 + c3 - abc - abc - abc 

Join like terms i.e (-abc) and we get: 
= a3 + b3 + c3 - 3abc 

Hence, in this way we obtain the identity i.e. a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca) 

Answer 2

A,b, c are real numbers such that a+b+c=7, a^2+b^2+c^2=35,a^3+b^3+c^3=151. What is the value of a×b×c?

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2 ANSWERS



Aditya Pradeep, Likes to solve!!

Answered Jan 15, 2016

Solution: abc=-15

As ,there are 3 equations and 3 variables,there is a unique solution.

Approach 1: Lazy approach

I dont want to solve the equations, I see that sum of 3 cubes is 151. I guess that one of them is 5.

Hence by some trial and error we see that 5,3,-1 satisfies .

Approach 2: Proper algebra

Whenever we are given a problems like this it is better to construct a cubic equation whose roots are a,b,c and solve the cubic.

So we need a+b+c, ab+bc+ca and abc.

a+b+c =7 ( given)

Squaring and expanding both sides,

a2+b2+c2+2(ab+bc+ca)=49a2+b2+c2+2(ab+bc+ca)=49

hence ab+bc+ca=7ab+bc+ca=7

Lastly we use the identity

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

we know all terms except abc,

On solving abc=-15.

Which is what is asked.

However, I will go a bit further and determine the values of a,b and c.

We get a cubic

t3−7t2+7t+15=0t3−7t2+7t+15=0

(t−5)(t−3)(t+1)=0(t−5)(t−3)(t+1)=0

Hence -1,3 and 5 are values of abc.

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