NMTC 2015 junior Q1

Let $\sqrt[3]{x-1}$ be $a$ ; $\sqrt [3]{x-3}$ be $b$ ; and $\sqrt [3]{x-5}$ be $c$ .

Then we have $a+b+c =0\Rightarrow {a^3+b^3+c^3=3abc}$

Putting the respective values in it it,we get $x-1+x-3+x-5=3\sqrt [3]{(x-1)(x-3)(x-5)}$ $\Rightarrow {3x-9=3\sqrt [3]{(x-1)(x-3)(x-5)}}$ $\Rightarrow{x-3=\sqrt [3]{(x-1)(x-3)(x-5)}}$ Taking cube of both sides we get $x^3-9x^2+27x-27=x^3-9x^2+23x-15$ $\Rightarrow{27x-23x=27-15}\Rightarrow{4x=12}$ $\boxed{x=3}$ Therefore number of solution of the equation is $\huge {1}$ .

The LHS is an increasing continuous function $f(x)$ with $f(1)<0$ and $f(5)>0$ . Thus the equation $f(x)=0$ has exactly $\boxed{1}$ real solution.

x=3 is the only solution for the equation...... Hence the answer is 1!!!!!

Satyajit Ghosh write the solution I told you in the morning