Find the product of 4 root 27 + root 9 and 3 root 3 -2

Open in App

Solution

$(4 \sqrt{27} + \sqrt{9}) (3 \sqrt{3} - 2) = [4 \sqrt{3 \times 3 \times 3} + \sqrt{3 \times 3}] [3 \sqrt{3} - 2] = (12 \sqrt{3} + 3) (3 \sqrt{3} - 2) = 12 \sqrt{3} (3 \sqrt{3} - 2) + 3 (3 \sqrt{3} - 2) = 108 - 24 \sqrt{3} + 9 \sqrt{3} - 6 = 102 - 15 \sqrt{3}$

Suggest Corrections

Similar questions

Find the product of the cube root of 27 and the cube root of -27. [2 MARKS]

\sqrt{27 a^{3} b^{2} c^{4}} \times \sqrt[3]{128 a^{7} b^{9} c^{2}} \times \sqrt[6]{729 a b^{12} c^{2}}

^{'} ω^{'}

ω^{(\frac{1}{3} + \frac{2}{9} + \frac{4}{27} + . . . \infty)} + ω^{(\frac{1}{2} + \frac{3}{8} + \frac{9}{32} + . . . \infty)} =

Join BYJU'S Learning Program