MCQ
The equation whose roots are $\frac{1}{{3 + \sqrt 2 }}$and $\frac{1}{{3 - \sqrt 2 }}$ is
- ✓$7{x^2} - 6x + 1 = 0$
- B$6{x^2} - 7x + 1 = 0$
- C${x^2} - 6x + 7 = 0$
- D${x^2} - 7x + 6 = 0$
${x^2} - \left( {\frac{1}{{3 + \sqrt 2 }} + \frac{1}{{3 - \sqrt 2 }}} \right)x + \frac{1}{{3 + \sqrt 2 }} \times \frac{1}{{3 - \sqrt 2 }} = 0$
$ \Rightarrow {x^2} - \left( {\frac{6}{7}} \right)x + \frac{1}{7} = 0 \Rightarrow 7{x^2} - 6x + 1 = 0$
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Statement $-2$ : $\mathop \sum \limits_{k = 1}^n \left( {{k^3} - {{\left( {k - 1} \right)}^3}} \right) = {n^3},$ for any natural number $n$