Find cube of a 1 2 - b 1 5 - c 1 3 - d 2-

The correct option is D a ^ 3 2 b ^ 3 5 / c d ^ 6 ( a ^1 2 b ^ 1 5 / c ^ 1 3 d ^ 2 #160;) #160;^ 3 = a ^3 2 b ^ 3 5 / c d ^ 6 #160; ...

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Byju's Answer

Standard VIII

Mathematics

Cube-Root of Fractional Numbers

Find cube of ...

Question

Find cube of $\frac{a^{\frac{1}{2}} \times b^{\frac{1}{5}}}{c^{\frac{1}{3}} \times d^{2}}$ :

\frac{a^{\frac{3}{2}} b^{\frac{3}{5}}}{c d^{3}}

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\frac{a^{\frac{3}{2}} b^{\frac{1}{5}}}{c d^{6}}

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\frac{a^{\frac{3}{2}} b^{\frac{1}{5}}}{c^{\frac{1}{3}} d^{2}}

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\frac{a^{\frac{3}{2}} b^{\frac{3}{5}}}{c d^{6}}

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Solution

The correct option is D $\frac{a^{\frac{3}{2}} b^{\frac{3}{5}}}{c d^{6}}$
${⎛ ⎜ ⎝ \frac{a^{\frac{1}{2}} b^{\frac{1}{5}}}{c^{\frac{1}{3}} d^{2}} ⎞ ⎟ ⎠}^{3} = \frac{a^{\frac{3}{2}} b^{\frac{3}{5}}}{c d^{6}}$

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If A1, B1, C1 … are the cofactors of the elements a1, b1, c1... of the matrix

$⎡ ⎢ ⎣ \begin{matrix} a 1 & b 1 & c 1 a 2 & b 2 & c 2 a 3 & b 3 & c 3 \end{matrix} ⎤ ⎥ ⎦$ then $∣ ∣ ∣ \begin{matrix} B 2 & C 2 B 3 & C 3 \end{matrix} ∣ ∣ ∣ =$

Q. If

| a_{1} | > | a_{2} | + | a_{3} |, | b_{2} | > | b_{1} | + | b_{3} |

and

| c_{3} | > | c_{1} | + | c_{2} |

, then show

∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0

Q. If

(a_{1}, a_{2}, a_{3})

(b_{1}, b_{2}, b_{3})

and

(c_{1}, c_{2}, c_{3})

are linearly independent and

x (a_{1}, a_{2}, a_{3}) + y (b_{1}, b_{2}, b_{3}) + z (c_{1}, c_{2}, c_{3}) = 0

then

Q. If

2 y - x = 8; 5 y - x = 14, y - 2 x = 1

intersect at

(a_{1}, b_{1}), (a_{2}, b_{2}), (a_{3}, b_{3})

Find

(a_{1} + a_{2} + a_{3} + b_{1} + b_{2} + b_{3})

Q. Suppose four distinct positive number

a_{1}, a_{2}, a_{3}, a_{4}

are in G.P. Let

b_{1} = a_{1}, b_{2} = b_{1} + a_{2}, a_{3} = b_{2} + a_{3}

and

b_{4} = b_{3} + a_{4}

Statement 1 - The numbers

b_{1}, b_{2}, b_{3}, b_{4}

are neither in A.P. nor in G.P.
Statement 2 - The numaber

b_{1}, b_{2}, b_{3}, b_{4}

are in H.P.

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Find cube of a12×b15c13×d2:

The correct option is D a32b35cd6⎛⎜⎝a12b15c13d2⎞⎟⎠3=a32b35cd6

Find cube of $\frac{a^{\frac{1}{2}} \times b^{\frac{1}{5}}}{c^{\frac{1}{3}} \times d^{2}}$ :

The correct option is D $\frac{a^{\frac{3}{2}} b^{\frac{3}{5}}}{c d^{6}}$
${⎛ ⎜ ⎝ \frac{a^{\frac{1}{2}} b^{\frac{1}{5}}}{c^{\frac{1}{3}} d^{2}} ⎞ ⎟ ⎠}^{3} = \frac{a^{\frac{3}{2}} b^{\frac{3}{5}}}{c d^{6}}$