Question

solve the system of equation in natural numbers a^-b^-c^=3abc; a^=2(b+c)



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Answer 1

Answer:

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Answer 2

we know,

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

Now it is given that : a3 + b3 + c3 = 3abc 

So,

a3 + b3 + c3 - 3abc = 0 

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

this means either

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

(a2 + b2 + c2 -ab - ac -bc) = 0 cannot be zero because:

2a² + 2b² + 2c² -2ab - 2ac - 2bc = 0 

a² + b² -2ab + a² +b² +2c² - 2ac -2bc = 0 

(a-b)² + a² + c² -2ac + b² + c² -2bc = 0 

(a-b)² + (a-c)² + (b-c)² = 0

(a-b)² , (a-c)² ,(b-c)² >=0

As a≠b≠c , The given value cannot be zero.

This means (a + b + c ) has to be zero