Question : If $\mathrm{p}=7+4 \sqrt{3}$, then what is the value of $\frac{\mathrm{p}^6+\mathrm{p}^4+\mathrm{p}^2+1}{\mathrm{p}^3}$?
Option 1: 2617
Option 2: 2167
Option 3: 2716
Option 4: 2176
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: 2716
Solution : Given, $p=7+4\sqrt3$ ⇒ $\frac1p=\frac{1}{7+4\sqrt3}$ ⇒ $\frac1p=\frac{1}{7+4\sqrt3}\times\frac{7-4\sqrt3}{7-4\sqrt3}$ ⇒ $\frac{1}{p}=\frac{7-4\sqrt3}{(7+4\sqrt3)(7-4\sqrt3)}$ We know $(a+b)(a-b)=a^2-b^2$ ⇒ $\frac1p=\frac{7-4\sqrt3}{7^2-(4\sqrt3)^2}$ ⇒ $\frac1p=\frac{7-4\sqrt3}{49-48}$ ⇒ $\frac1p=7-4\sqrt3$ ⇒ $p+\frac1p=7+4\sqrt3+7-4\sqrt3=14$ Now, consider $\frac{p^6+p^4+p^2+1}{p^3}$ $=\frac{p^6+1}{p^3}+\frac{p^4+p^2}{p^3}$ $=(p^3+\frac{1}{p^3})+(p+\frac{1}{p})$ We know, $(a+b)^3=a^3+b^3+3ab(a+b)$ So, $=(p^3+\frac{1}{p^3})+(p+\frac{1}{p})=(p+\frac1p)^3-3(p+\frac{1}{p})+(p+\frac1p)$ $=(p+\frac1p)^3-2(p+\frac{1}{p})$ $=14^3-2\times14$ $=2744-28$ $=2716$ Hence, the correct answer is 2716.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $\mathrm{p}=\frac{\sqrt{2}+1}{\sqrt{2}-1}$ and $\mathrm{q}=\frac{\sqrt{2}-1}{\sqrt{2}+1}$ then, find the value of $\frac{\mathrm{p}^2}{\mathrm{q}}+\frac{\mathrm{q}^2}{\mathrm{p}}$.
Question : If $7\mathrm{b}-\frac{1}{4 \mathrm{b}}=7$, what is the value of $16 \mathrm{b}^2+\frac{1}{49 \mathrm{b}^2}$?
Question : If $\sqrt{1+\frac{\sqrt{3}}{2}}-\sqrt{1-\frac{\sqrt{3}}{2}}=c$, then the value of $\mathrm{c}$ is:
Question : If $\frac{\sqrt{26-7 \sqrt{3}}}{\sqrt{14+5 \sqrt{3}}}=\frac{b+a \sqrt{3}}{11}, b>0$, then what is the value of $\sqrt{(\mathrm{b}-\mathrm{a})}$?
Question : If $\sec ^2 \mathrm{~A}+\tan ^2 \mathrm{~A}=3$, then what is the value of $\cot \mathrm{A}$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile