Question
The value of $(3 +2√2)^{-3}+ (3 -2√2)^{- 3}$ is
Answer: Option A
Answer:(d)$(3 +2√2)^{-3}+ (3 -2√2)^{- 3}$$(3 + 2√2) (3 - 2√ 2)$= $(3)^2 - (2√2)^2$ = 9 - 8 = 1$3 + 2√2 = 1/{3 - 2√2}$$(x + y)^3 + (x - y)^3 = x^3 + y^3 + 3x^2 y + 3 x y^2 + x^3 - y^3 - 3 x^2 y + 3 x y^2$= $2 x^3 + 6 x y^2$$(3 + 2√2 )^{ - 3} + (3 - 2√2)^{ - 3}$= $(1/{3 + 2√2})^3 +(1/{3 -2√2})^3$= $(3 - 2√2)^3 + (3 + 2√2)^3$= $2 × (3)^3 + 6 × 3 × (2√2)^2$= 2 × 27 + 18 × 8= 54 + 144 = 198
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Answer:(d)$(3 +2√2)^{-3}+ (3 -2√2)^{- 3}$$(3 + 2√2) (3 - 2√ 2)$= $(3)^2 - (2√2)^2$ = 9 - 8 = 1$3 + 2√2 = 1/{3 - 2√2}$$(x + y)^3 + (x - y)^3 = x^3 + y^3 + 3x^2 y + 3 x y^2 + x^3 - y^3 - 3 x^2 y + 3 x y^2$= $2 x^3 + 6 x y^2$$(3 + 2√2 )^{ - 3} + (3 - 2√2)^{ - 3}$= $(1/{3 + 2√2})^3 +(1/{3 -2√2})^3$= $(3 - 2√2)^3 + (3 + 2√2)^3$= $2 × (3)^3 + 6 × 3 × (2√2)^2$= 2 × 27 + 18 × 8= 54 + 144 = 198
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