Let \(\displaystyle{c}_{{1}},{c}_{{2}},{c}_{{3}}\) be the constants.

So \(\displaystyle{u}{c}_{{1}}+{v}{c}_{{2}}+{w}{c}_{{3}}={0}\) gives

\(\displaystyle{c}_{{1}}{\left({t}^{{3}}-{4}{t}^{{2}}+{3}{t}+{3}\right)}+{c}_{{2}}{\left({t}^{{3}}+{2}{t}^{{2}}+{4}{t}-{1}\right)}+{c}_{{3}}{\left({2}{t}^{{3}}-{t}^{{2}}={3}{t}+{5}\right)}={0}\)

\(\displaystyle{t}^{{3}}{\left({c}_{{1}}+{c}_{{2}}+{2}{c}_{{3}}\right)}+{t}^{{2}}{\left(-{4}{c}_{{1}}+{2}{c}_{{2}}-{c}_{{3}}\right)}+{t}{\left({3}{c}_{{1}}+{4}{c}_{{2}}-{3}{c}_{{3}}\right)}+{\left({3}{x}_{{1}}-{c}_{{2}}+{5}{c}_{{3}}\right)}={0}\)

Comparing coefficients of \(\displaystyle{t}^{{3}},{t}^{{2}},{t}\) and constant term both sides we get,

\(\displaystyle{c}_{{1}}+{c}_{{2}}+{2}{c}_{{3}}={0}\) (1)

\(\displaystyle-{4}{c}_{{1}}+{2}{c}_{{2}}-{c}_{{3}}={0}\) (2)

\(\displaystyle{3}{c}_{{1}}+{4}{c}_{{2}}-{3}{c}_{{3}}=\) (3)

and \(\displaystyle{3}{c}_{{1}}-{c}_{{2}}+{5}{c}_{{3}}={0}\) (4)

Subtracting equation (4) from equation (3), we get

\(\displaystyle{8}{c}_{{3}}={5}{c}_{{2}}\)

\(\displaystyle{c}_{{3}}={\left(\frac{{5}}{{8}}\right)}{c}_{{2}}\) (5)

Substituting the value of \(c_3\) in equation (1), we get

\(\displaystyle{c}_{{1}}+{c}_{{2}}+{2}{\left(\frac{{5}}{{8}}\right)}{c}_{{2}}={0}\)

\(\displaystyle{c}_{{1}}={\left(-\frac{{9}}{{4}}\right)}{c}_{{2}}\) (6)

Substituting values from (5) & (6) in equation (1), we get

\(\displaystyle{\left(-\frac{{9}}{{4}}\right)}{c}_{{2}}+{c}_{{2}}+{\left(\frac{{5}}{{8}}\right)}{c}_{{2}}={0}\)

So, \(\displaystyle{c}_{{2}}={0}\)

And so from (5) and (6) \(\displaystyle{c}_{{1}}={c}_{{2}}={c}_{{3}}={0}\)

Therefore, polynomials \(\displaystyle{u}={t}^{{3}}-{4}{t}^{{2}}+{3}{t}+{3},{v}={t}^{{3}}+{2}{t}^{{2}}+{4}{t}-{1},{w}={2}{t}^{{3}}-{t}^{{2}}-{3}{t}+{5}\) are linearly independent.