The value of $${\left( {3 + 2\sqrt 2 } \right)^{ - 3}}$$ $$ + {\left( {3 - 2\sqrt 2 } \right)^{ - 3}} = ?$$
A. 189
B. 180
C. 108
D. 198
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & {\left( {3 + 2\sqrt 2 } \right)^{ - 3}} + {\left( {3 - 2\sqrt 2 } \right)^{ - 3}} \cr & = {\left( {\frac{1}{{3 + 2\sqrt 2 }}} \right)^3} + {\left( {\frac{1}{{3 - 2\sqrt 2 }}} \right)^3} \cr} $$$$ = {\left( {\frac{1}{{\left( {3 - 2\sqrt 2 } \right)}} \times \frac{{3 + 2\sqrt 2 }}{{3 + 2\sqrt 2 }}} \right)^3} + $$ $${\left( {\frac{1}{{\left( {3 + 2\sqrt 2 } \right)}} \times \frac{{3 - 2\sqrt 2 }}{{3 - 2\sqrt 2 }}} \right)^3}$$
$$\eqalign{ & = {\left( {\frac{{3 - 2\sqrt 2 }}{{9 - 8}}} \right)^3} + {\left( {\frac{{3 + 2\sqrt 2 }}{{9 - 8}}} \right)^3} \cr & = {\left( {3 - 2\sqrt 2 } \right)^3} + {\left( {3 + 2\sqrt 2 } \right)^3} \cr & a = 3 - 2\sqrt 2 \cr & b = 3 + 2\sqrt 2 \cr & \left[ {\because {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)} \right] \cr & = \left( {3 - 2\sqrt 2 + 3 + 2\sqrt 2 } \right)\left( {17 + 17 - 1} \right) \cr & = \left( 6 \right)\left( {33} \right) \cr & = 198 \cr} $$
Related Questions on Surds and Indices
A. $$\frac{1}{2}$$
B. 1
C. 2
D. $$\frac{7}{2}$$
Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to:
A. 1.45
B. 1.88
C. 2.9
D. 3.7
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