number of valuesof $a$ for which vectors are coplanar what are the number of values of $a$ for which $-a^2i+j+k,i-a^2j+k,i+j-a^2k$ are coplanar. $$\text{attempt}$$ For vectors to be coplanar the determinant of their coefficients should be zero. thus what i get after expanding is $a^6-3a^2-2=0$. Now I dont know how many roots of this polynomial are real values.putting $a^2=x$ doesnt simplify much as we still have to deal with cubic. Thank you!
 A: You have a good start. Substituting $a^2=x$, transforms this equation into
$$x^3-3x-2=0.$$
Although it is a cubic equation, this one is fortunately not as difficult as they can be. We can try to guess a simple root of this equation. Of course, polynomials don't have to have integer roots; but if a polynomial with integer coefficients has an integer root, then it has to be a divisor of the constant term. Checking divisors of the constant term $-2$, i.e. numbers $\pm1,\pm2$, we can see that $x=2$ is a root. Therefore the polynomial on the left-hand side is divisible by $(x-2)$. Using either long or synthetic division, this polynomial factors into
$$x^3-3x-2=(x-2)(x^2+2x+1)=(x-2)(x+1)^2=0.$$
So the roots of this equation are $x=2$ and $x=-1$. I hope you can take it from here and finish solving for $a$.
Note: during the trial-and-error step, I just happened to stumble upon the root $x=2$ first. Of course, we could've found $x=-1$ first, but then — i.e. after we found any one root — the procedure would be the same.
A: The eigenvalues of the coefficient matrix are easily found to be $2-a^2$ and $-(1+a^2)$ (see here for a brief explanation of how to do this), which are also the factors of the determinant. For the determinant to be zero, at least one eigenvalue must be, so we either have $2-a^2=0$ or $1+a^2=0$. Therefore, if $a$ is restricted to real numbers, $a=\pm\sqrt2$. 
