Question

Find the value of $\left(2^{\frac{1}{4}}-1\right)\left(2^{\frac{3}{4}}+2^{\frac{1}{2}}+2^{\frac{1}{4}}+1\right).$

• A. 2
• B. -1
• C. -2
• D. 1

Answer 1

The correct option is D

1

$\begin{array}{c}Putting{2}^{\frac{1}{4}}=x,weget:\\ \left(2^{\frac{1}{4}}-1\right)\left(2^{\frac{3}{4}}+2^{\frac{1}{2}}+2^{\frac{1}{4}}+1\right)=\left(x-1\right)\left(x^3+x^2+x+1\right),wherex=2^{\frac{1}{4}}\\ =\left(x-1\right)\left[x^2\left(x+1\right)+\left(x+1\right)\right]\\ =\left(x-1\right)\left(x+1\right)\left(x^2+1\right)=\left(x^2-1\right)\left(x^2+1\right)\\ =\left(x^4-1\right)=\left[{\left(2^{\frac{1}{4}}\right)}^4-1\right]=\left[2^{\left(\frac{1}{4}\times 4\right)}-1\right]=\left(2-1\right)=1.\end{array}$