Check if polynomials are linearly independent.

Just for fun you could use the definition of linear independence. That is, say that $$ a(1) + b(1 + t^2) + c(1 + t + t^2) = 0. $$ Then $$ (a + b + c)1 + (c)t + (b + c)t^2 = 0. $$ Now you probably know that $\{1,t,t^2\}$ is linearly independent. So $$ \begin{align} c &= 0 \\ b + c &= 0 \Rightarrow b = 0 \\ a + b + c &= 0 \Rightarrow a = 0. \end{align} $$

The coordinate vectors with respect to the basis $\{1,t,t^2\}$ are $$ \begin{bmatrix}1\\0\\0\end{bmatrix} \quad \begin{bmatrix}1\\0\\1\end{bmatrix} \quad \begin{bmatrix}1\\1\\1\end{bmatrix} $$ so the matrix you have to compute the rank of is $$ \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix} $$ Swapping rows 2 and 3 brings it into unitriangular form, so the rank is $3$.

Alternatively you can observe that $$ t=(1+t+t^2)-(1+t^2)\\ t^2=(1+t^2)-1 $$