The number of distinct real roots of the equation $x^7-7x-2=0$ is

$f\left(x\right)=x^7-7x-2$
$f^{\text{'}}\left(x\right)=7x^6-7$
$=7\left(x^6-1\right)$
$=7\left(x^3+1\right)\left(x^3-1\right)$
$f^{\text{'}}\left(x\right)=7(x+1)(x-1)\left(x^2+x+1\right)\left(x^2-x+1\right)$
$f^{\text{'}}\left(x\right)=0$ has two real roots
$f\left(1\right)=-8$
f(-1)=4

f(2)=112
f(-2)=-116
f(0)=-2

The graphs of y=x⁷-7x and y=2  intersect at 3 points 

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So by graph f(x)=0 has three real roots