If $a^{2}$ is a perfect square and $b^{3}$ is a perfect cube, then $a^{2} \times b^{3}$ is:

perfect square

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perfect cube

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either of

A

B

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neither

A

nor

B

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Solution

The correct option is D neither $A$ nor $B$
$a^{2} \times b^{3} = a^{2} b^{3}$
It is not a perfect square as we need $1$ more $b$
It is not perfect cube as we need $1$ more $a$ .

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(ii) 8640 is not a perfect cube.
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(iv) There is no perfect cube which ends in 4.
(v) For an integer a, a³ is always greater than a².
(vi) If a and b are integers such that a² > b², then a³ > b³.
(vii) If a divides b, then a³ divides b³.
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A^{3} = B^{3}

and A(AB) = B(BA) , then

Q. If

c = a^{3} - b^{3},

then is

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a perfect cube?

Let A={a1,a2,a3,a4,a5} and B={b1,b2,b3,b4,} where ai's and bi's are school going students . Define a relation from A to B by xRy iff y is a true friend of x . If R={(a1,b1),(a2,b1),(a3,b3),(a4,b2},(a5,b2)} . Prove that R is neither one one nor onto

Q. If

x = \sqrt[3]{a + \sqrt{a^{2} - b^{3}}} + \sqrt[3]{a - \sqrt{a^{2} - b^{3}}}

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If a2 is a perfect square and b3 is a perfect cube, then a2×b3 is:

The correct option is D neither A nor Ba2×b3=a2b3It is not a perfect square as we need 1 more bIt is not perfect cube as we need 1 more a.

If $a^{2}$ is a perfect square and $b^{3}$ is a perfect cube, then $a^{2} \times b^{3}$ is:

The correct option is D neither $A$ nor $B$
$a^{2} \times b^{3} = a^{2} b^{3}$
It is not a perfect square as we need $1$ more $b$
It is not perfect cube as we need $1$ more $a$ .