Find the cube of 2 a -1-2 a q-0-A- 4 a3-1-4 a3-4 a-8-2 aB- 8 a 3-1-8 a 3-6 a -3-2 aC- 8 a3-1-8 a3-6 a-3-2 aD- 16 a 3-1-8 a3-6 a -8-2 a

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 IX

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

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

Find the cube of $(2 a - 6 b)$

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)

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

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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) .