Find the cube of 2a-1-2a- - a -0- -4a-3 - 1-4a-3- 4a - 8-2a8a-3 - 1-8a-3 - 6a - 3-2a8a-3 - 1-8a-3- 6a - 3-2a16a-3 - 1-8a-3- 6a - 8-2a

The correct option is C 8a^3 + 1/8a^3 + 6a + 3/2a( a + b )^3 = a^3 + b^3 + 3ab(a + b) (2a+1/2a)^3 = (2a)^3 + (1/2a)^3 + 3×2a×1/2a(2a+1/2a) = 8a^3 + 1/8a^3 + ...

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Byju's Answer

Standard VIII

Mathematics

Expansion of (a+b)^3

Find the cube...

Question

Find the cube of $2 a + \frac{1}{2 a}$ ( $a \neq$ 0) .

4 a^{3} + \frac{1}{4 a^{3}} + 4 a + \frac{8}{2 a}

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8 a^{3} - \frac{1}{8 a^{3}} - 6 a - \frac{3}{2 a}

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8 a^{3} + \frac{1}{8 a^{3}} + 6 a + \frac{3}{2 a}

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16 a^{3} - \frac{1}{8 a^{3}} + 6 a - \frac{8}{2 a}

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Solution

The correct option is C $8 a^{3} + \frac{1}{8 a^{3}} + 6 a + \frac{3}{2 a}$
$(a + b)^{3} = a^{3} + b^{3} + 3 a b (a + b)$

$(2 a + \frac{1}{2 a})^{3} = (2 a)^{3} + (\frac{1}{2 a})^{3} + 3 \times 2 a \times \frac{1}{2 a} (2 a + \frac{1}{2 a})$

= $8 a^{3} + \frac{1}{8 a^{3}} + 3 (2 a + \frac{1}{2 a})$

= $8 a^{3} + \frac{1}{8 a^{3}} + 6 a + \frac{3}{2 a}$

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Similar questions

The factors of $8 a^{3} + b^{3} - 6 a b + 1$ are:

Q. The factors of 8a³ + b³ − 6ab + 1 are

(a) (2a + b − 1) (4a² + b² + 1 − 3ab − 2a)

(b) (2a − b + 1) (4a² + b² − 4ab + 1 − 2a + b)

(c) (2a + b + 1) (4a² + b² + 1 −2ab − b − 2a)

(d) (2a − 1 + b) (4a² + 1 − 4a − b − 2ab)

Q. Find the cube of

2 a + \frac{1}{2 a}

(

a \neq

0) .

Q. State whether the statement is True or False.

Expand:

(2 a - \frac{1}{2 a})^{3}

is equal to

8 a^{3} - 6 a + \frac{3}{2 a} - \frac{1}{8 a^{3}}

Q. State whether the statement is True or False.

The cube of

(2 a - 1)

is equal to

8 a^{3} - 12 a^{2} + 6 a - 1

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Find the cube of 2a+12a ( a≠0) .

The correct option is C 8a3+18a3+6a+32a(a+b)3=a3+b3+3ab(a+b) (2a+12a)3=(2a)3+(12a)3+3×2a×12a(2a+12a) = 8a3+18a3+3(2a+12a) = 8a3+18a3+6a+32a

Find the cube of $2 a + \frac{1}{2 a}$ ( $a \neq$ 0) .