To the power negative?

Algebra Level 2

If a = ( 3 + 2 ) − 3 a=(\sqrt{3}+\sqrt{2})^{-3} and b = ( 3 − 2 ) − 3 b=(\sqrt{3}-\sqrt{2})^{-3} , find the value of ( a + 1 ) − 1 + ( b + 1 ) − 1 (a+1)^{-1}+(b+1)^{-1} .


The answer is 1.

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Ayush G Rai by Ayush G Rai , 20, India

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2 solutions

Sabhrant Sachan
May 11, 2016

Relevant wiki: Surds

We have to calculate : ( a + 1 ) − 1 + ( b + 1 ) − 1 ⟹ 1 a + 1 + 1 b + 1 ⟹ a + b + 2 a b + a + b + 1 a b = ( 3 + 2 ) − 3 × ( 3 − 2 ) − 3 = 1 ⟹ a + b + 2 a b + a + b + 1 = a + b + 2 a + b + 2 ⟹ A n s = 1 \text{We have to calculate : }(a+1)^{-1}+(b+1)^{-1} \implies \dfrac{1}{a+1}+\dfrac{1}{b+1} \implies \dfrac{a+b+2}{ab+a+b+1} \\ ab=(\sqrt{3}+\sqrt{2})^{-3}\times(\sqrt{3}-\sqrt{2})^{-3}=1 \\ \implies\dfrac{a+b+2}{ab+a+b+1} = \dfrac{a+b+2}{a+b+2} \implies \boxed{Ans=1}

9 Helpful 0 Interesting 0 Brilliant 0 Confused

very goooood solution...+1

Ayush G Rai - 5 years, 1 month ago

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thank you ayush ...

Sabhrant Sachan - 5 years, 1 month ago

@Sabhrant Sachan How do you know that a b = ( 3 + 2 ) − 3 × ( 3 − 2 ) − 3 = 1 \displaystyle ab = ( \sqrt { 3 } + \sqrt { 2 } ) ^ { -3 } \times ( \sqrt { 3 } - \sqrt { 2 } ) ^ { -3 } = 1 ?

. . - 3 months ago

( 3 + 2 ) − 3 × ( 3 − 2 ) − 3 = ( 3 − 2 ) − 3 = 1 (\sqrt{3}+\sqrt{2})^{-3}\times(\sqrt{3}-\sqrt{2})^{-3}= (3-2)^{-3} = 1

Sabhrant Sachan - 1 month, 3 weeks ago
Hung Woei Neoh
May 11, 2016

a = ( 3 + 2 ) − 3 = 1 ( 3 + 2 ) 3 a + 1 = 1 ( 3 + 2 ) 3 ‘ + 1 = 1 + ( 3 + 2 ) 3 ( 3 + 2 ) 3 ( a + 1 ) − 1 = ( 1 + ( 3 + 2 ) 3 ( 3 + 2 ) 3 ) − 1 = ( 3 + 2 ) 3 1 + ( 3 + 2 ) 3 b = ( 3 − 2 ) − 3 = 1 ( 3 − 2 ) 3 b + 1 = 1 ( 3 − 2 ) 3 ‘ + 1 = 1 + ( 3 − 2 ) 3 ( 3 − 2 ) 3 ( b + 1 ) − 1 = ( 1 + ( 3 − 2 ) 3 ( 3 − 2 ) 3 ) − 1 = ( 3 − 2 ) 3 1 + ( 3 − 2 ) 3 a=(\sqrt{3}+\sqrt{2})^{-3}=\dfrac{1}{(\sqrt{3}+\sqrt{2})^3}\\ a+1 = \dfrac{1}{(\sqrt{3}+\sqrt{2})^3}`+ 1 = \dfrac{1 + (\sqrt{3}+\sqrt{2})^3}{(\sqrt{3}+\sqrt{2})^3}\\ (a+1)^{-1} = \left(\dfrac{1 + (\sqrt{3}+\sqrt{2})^3}{(\sqrt{3}+\sqrt{2})^3}\right)^{-1} = \dfrac{(\sqrt{3}+\sqrt{2})^3}{1 + (\sqrt{3}+\sqrt{2})^3}\\ b=(\sqrt{3}-\sqrt{2})^{-3}=\dfrac{1}{(\sqrt{3}-\sqrt{2})^3}\\ b+1 = \dfrac{1}{(\sqrt{3}-\sqrt{2})^3}`+ 1 = \dfrac{1 + (\sqrt{3}-\sqrt{2})^3}{(\sqrt{3}-\sqrt{2})^3}\\ (b+1)^{-1} = \left(\dfrac{1 + (\sqrt{3}-\sqrt{2})^3}{(\sqrt{3}-\sqrt{2})^3}\right)^{-1} = \dfrac{(\sqrt{3}-\sqrt{2})^3}{1 + (\sqrt{3}-\sqrt{2})^3}

( a + 1 ) − 1 + ( b + 1 ) − 1 = ( 3 + 2 ) 3 1 + ( 3 + 2 ) 3 + ( 3 − 2 ) 3 1 + ( 3 − 2 ) 3 = ( 3 + 2 ) 3 ( 1 + ( 3 − 2 ) 3 ) + ( 3 − 2 ) 3 ( 1 + ( 3 + 2 ) 3 ) ( 1 + ( 3 − 2 ) 3 ) ( 1 + ( 3 + 2 ) 3 ) = ( 3 + 2 ) 3 + ( 3 − 2 ) 3 + 2 ( 3 + 2 ) 3 ( 3 − 2 ) 3 1 + ( 3 + 2 ) 3 + ( 3 − 2 ) 3 + ( 3 + 2 ) 3 ( 3 − 2 ) 3 = ( 3 + 2 ) 3 + ( 3 − 2 ) 3 + 2 ( ( 3 + 2 ) ( 3 − 2 ) ) 3 1 + ( 3 + 2 ) 3 + ( 3 − 2 ) 3 + ( ( 3 + 2 ) ( 3 − 2 ) ) 3 = ( 3 + 2 ) 3 + ( 3 − 2 ) 3 + 2 ( 3 − 2 ) 3 1 + ( 3 + 2 ) 3 + ( 3 − 2 ) 3 + ( 3 − 2 ) 3 = ( 3 + 2 ) 3 + ( 3 − 2 ) 3 + 2 ( 1 ) 3 1 + ( 3 + 2 ) 3 + ( 3 − 2 ) 3 + ( 1 ) 3 = ( 3 + 2 ) 3 + ( 3 − 2 ) 3 + 2 ( 3 + 2 ) 3 + ( 3 − 2 ) 3 + 2 = 1 (a+1)^{-1} + (b+1)^{-1}\\ =\dfrac{(\sqrt{3}+\sqrt{2})^3}{\color{#EC7300}{1 + (\sqrt{3}+\sqrt{2})^3}} + \dfrac{(\sqrt{3}-\sqrt{2})^3}{\color{#20A900}{1 + (\sqrt{3}-\sqrt{2})^3}}\\ =\dfrac{(\sqrt{3}+\sqrt{2})^3 \color{#20A900}{\left(1 + (\sqrt{3}-\sqrt{2})^3 \right)} +(\sqrt{3}-\sqrt{2})^3 \color{#EC7300}{\left(1 + (\sqrt{3}+\sqrt{2})^3 \right)}}{\color{#20A900}{\left(1 + (\sqrt{3}-\sqrt{2})^3 \right)} \color{#EC7300}{\left(1 + (\sqrt{3}+\sqrt{2})^3 \right)}}\\ =\dfrac{(\sqrt{3}+\sqrt{2})^3 + (\sqrt{3}-\sqrt{2})^3 +2\color{#D61F06}{(\sqrt{3}+\sqrt{2})^3 (\sqrt{3}-\sqrt{2})^3}}{1 +(\sqrt{3}+\sqrt{2})^3 + (\sqrt{3}-\sqrt{2})^3 +\color{#D61F06}{(\sqrt{3}+\sqrt{2})^3 (\sqrt{3}-\sqrt{2})^3}}\\ =\dfrac{(\sqrt{3}+\sqrt{2})^3 + (\sqrt{3}-\sqrt{2})^3 +2\color{#D61F06}{\left((\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})\right)^3}}{1 +(\sqrt{3}+\sqrt{2})^3 + (\sqrt{3}-\sqrt{2})^3 +\color{#D61F06}{\left((\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})\right)^3}}\\ =\dfrac{(\sqrt{3}+\sqrt{2})^3 + (\sqrt{3}-\sqrt{2})^3 +2\color{#D61F06}{\left(3-2\right)^3}}{1 +(\sqrt{3}+\sqrt{2})^3 + (\sqrt{3}-\sqrt{2})^3 +\color{#D61F06}{\left(3-2\right)^3}}\\ =\dfrac{(\sqrt{3}+\sqrt{2})^3 + (\sqrt{3}-\sqrt{2})^3 +2\left(1\right)^3}{1 +(\sqrt{3}+\sqrt{2})^3 + (\sqrt{3}-\sqrt{2})^3 +\left(1\right)^3}\\ =\dfrac{(\sqrt{3}+\sqrt{2})^3 + (\sqrt{3}-\sqrt{2})^3 +2}{(\sqrt{3}+\sqrt{2})^3 + (\sqrt{3}-\sqrt{2})^3 +2}\\ =\boxed{1}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

very lengthy solution but good..+1

Ayush G Rai - 5 years, 1 month ago

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