Question : If $x=(\sqrt[3]{7})^{3}+3$, then the value of $x^3–9x^2+27x–34$ is:
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: –1
Correct Answer: 0
Solution : Given: $(x–3)^{3}=(\sqrt[3]{7})^{3}$ $x^{3}–27–3(3x)(x–3) = 7$ $x^{3}–27–9x^{2}+27x = 7$ Subtracting 7 on both sides, we get, $x^{3}–27–9x^{2}+27x–7= 7–7$ $x^{3}–9x^{2}+27x–34 = 0$ Hence, the correct answer is 0.
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Question : If $a= \frac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-2}}$, then the value of $(a^{2}-ax)$ is:
Question : If $x^2-\sqrt{7} x+1=0$, then $\left(x^3+x^{-3}\right)=?$
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$; then what is the value of $x+\frac{1}{x}$?
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$, what is the value of $(x^3+\frac{1}{x^3})$?
Question : If $x^2-9x+1=0$, what is the value of $(x^3+\frac{1}{x^3})$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile