Find the cube of 2a-1-2a a -0 -

( 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 + 3(2a+1/2a) = 8a^3 + 1/8a^3 nbsp;+ 6a + 3/2a ...

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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) .

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

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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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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

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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) .