The value of -a1-8-a-1-8a1-8-a-1-8a1-4-a-1-4a1-2-a-1-2 is -A- a-a-1B- a-a-1- a2-a-2D- a1-2-a-1

The correct option is B Using (a + b) (a b) = a^2 b^2,we get : (a^1/8 + a^ 1/8) (a^1/8 a^ 1/8) (a^1/4 + a^ 1/4) (a^1/2 + a^ 1/2) =(a^1/4 a^ 1/4) (a^1 ...

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Byju's Answer

Standard IX

Mathematics

Surds

The value of ...

Question

The value of : $(a^{\frac{1}{8}} + a^{\frac{- 1}{8}}) (a^{\frac{1}{8}} - a^{\frac{- 1}{8}}) (a^{\frac{1}{4}} + a^{\frac{- 1}{4}}) (a^{\frac{1}{2}} + a^{\frac{- 1}{2}})$ is -

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Solution

The correct option is B

Using $(a + b) (a - b) = a^{2} - b^{2}$ ,we get :

$(a^{\frac{1}{8}} + a^{\frac{- 1}{8}}) (a^{\frac{1}{8}} - a^{\frac{- 1}{8}}) (a^{\frac{1}{4}} + a^{\frac{- 1}{4}}) (a^{\frac{1}{2}} + a^{\frac{- 1}{2}})$

= $(a^{\frac{1}{4}} - a^{\frac{- 1}{4}}) (a^{\frac{1}{4}} + a^{\frac{- 1}{4}}) (a^{\frac{1}{2}} + a^{\frac{- 1}{2}})$

= $(a^{\frac{1}{2}} - a^{\frac{- 1}{2}}) (a^{\frac{1}{2}} + a^{\frac{- 1}{2}}) = (a^{1} - a^{- 1})$ .

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Similar questions

Q. If positive square root of

a^{\frac{1}{a}} . (2 a)^{\frac{1}{2 a}} . (4 a)^{\frac{1}{4 a}} . (8 a)^{\frac{1}{8 a}} \dots \infty

\frac{8}{27},

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Q. If positive square root of

a^{\frac{1}{a}} . (2 a)^{\frac{1}{2 a}} . (4 a)^{\frac{1}{4 a}} . (8 a)^{\frac{1}{8 a}} \dots \infty

\frac{8}{27},

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lim x \to a {⎧ ⎨ ⎩ ⎡ ⎣ {(\frac{a^{1 / 2} + x^{1 / 2}}{a^{1 / 4} - x^{1 / 4}})}^{- 1} - \frac{2 (a x)^{1 / 4}}{x^{3 / 4} a^{1 / 4} x^{1 / 2} + a^{1 / 2} x^{1 / 4}} - a^{3 / 4} - \sqrt{2^{l o g_{4} a}} ⎤ ⎦ ⎫ ⎬ ⎭}^{8}

Q. If

a_{1} = \frac{1}{2 \times 5}, a_{2} = \frac{1}{5 \times 8}, a_{3} = \frac{1}{8 \times 11}, . . .,

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The value of : (a18+a−18)(a18−a−18)(a14+a−14)(a12+a−12) is -

The correct option is B Using (a+b)(a−b)=a2−b2,we get : (a18+a−18)(a18−a−18)(a14+a−14)(a12+a−12) =(a14−a−14)(a14+a−14)(a12+a−12) = (a12−a−12)(a12+a−12)=(a1−a−1).

The value of : $(a^{\frac{1}{8}} + a^{\frac{- 1}{8}}) (a^{\frac{1}{8}} - a^{\frac{- 1}{8}}) (a^{\frac{1}{4}} + a^{\frac{- 1}{4}}) (a^{\frac{1}{2}} + a^{\frac{- 1}{2}})$ is -