Question : If $\alpha$ and $\beta$ are the roots of equation $x^{2}-x+1=0$, then which equation will have roots $\alpha ^{3}$ and $\beta ^{3}?$
Option 1: $x^{2}+2x+1=0$
Option 2: $x^{2}-2x-1=0$
Option 3: $x^{2}+3x-1=0$
Option 4: $x^{2}-3x+1=0$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $x^{2}+2x+1=0$
Solution : $\alpha$ and $\beta$ are roots of the equation $x^{2}-x+1=0$ ⇒ $\alpha+\beta=1$ and $\alpha\beta=1$ Now we need to find the equation whose roots are $\alpha ^{3}$ and $\beta ^{3}$. ⇒ $(\alpha^{3}+\beta^{3})=(\alpha+\beta)^{3}-3\alpha\beta\times(\alpha+\beta)$ ⇒ $(\alpha^{3}+\beta^{3})=(1)^{3}-3\times 1 \times (1)$ ⇒ Sum of roots = $(\alpha^{3}+\beta^{3})=-2$ Also, Product of roots = $(\alpha^{3}\beta^{3})=(\alpha\beta)^{3} =1$ So, the required equation is $x^{2}+2x+1=0$ Hence, the correct answer is $x^{2}+2x+1=0$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $\alpha$ and $\beta$ are the roots of equation $x^{2}-2x+4=0$, then what is the equation whose roots are $\frac{\alpha ^{3}}{\beta ^{2}}$ and $\frac{\beta ^{3}}{\alpha ^{2}}?$
Question : If $\alpha, \beta$ are the roots of $6 x^2+13 x+7=0$, then the equation whose roots are $\alpha^2, \beta^2$ is:
Question : If $\tan (\alpha+\beta)=\sqrt{3}, \tan (\alpha-\beta)=1$ where $(\alpha+\beta)$ and $(\alpha-\beta)$ are acute angles, then what is $\tan$ $(6 \alpha) ?$
Question : If the sum of the roots of a quadratic equation is 1 and the product of the roots is –20, find the quadratic equation.
Question : If $\alpha +\beta =90^\circ$ and $\alpha :\beta =2:1,$ then the ratio of $\cos \alpha$ to $\cos \beta$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile