Question

A ,b,c three distinct real numbers then the num of distinct real roots of (x-a)^3+(x-b)^3+(x-c)^3=0 is

Answer 1

Hey There,
This question involves some small concepts of Polynomials, Limits and Application of Derivatives


Let's start: 
Let $f(x) = (x-a)^3+(x-b)^3+(x-c)^3$  
f(x) is a polynomial. And as properties of polynomials go: 

Here, We see that when $x \to -\infty$ then $f(x) \to -\infty$ 

And also, as $x \to \infty$ then $f(x) \to \infty$ 
This means that f(x) starts from negative infintity and goes till positive infintity. So, f(x) must have at least one real root, since it must intersect X-axis at at-least one point.  
Now,  
$f(x) = (x-a)^3+(x-b)^3+(x-c)^3 \\ \\ \implies f'(x) = 3(x-a)^2+3(x-b)^2+3(x-c)^2$

This shows that f'(x) is always positive.

Also, as a, b and c are three distinct numbers, f'(x) will also never be zero. 
This means that f(x) is a strictly increasing function. So, it can intersect the X-axis at at most one point.  This means, that f(x) has at most one real root .

Thus, finally we can come to the conclusion :  
f(x) has exactly one real root  .


Hope this helps
Purva
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