If one of the two factors of an expression which is the difference of two cubes is ( $x^{4}$ + $x^{2}$ y + $y^{2}$ ), then the other factor would be -

x + y

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x – y

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x² + y

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x² – y

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Solution

The correct option is D
x² – y

$x^{4}$ + $x^{2}$ y + $y^{2}$ = ${(x^{2})}^{2}$ + $x^{2}$ y + ${(y)}^{2}$

This is in the form of $a^{2}$ + ab + $b^{2}$ where a = $x^{2}$ , b = y

The other factor is a – b as $a^{3}$ – $b^{3}$ = (a - b)( $a^{2}$ + ab + $b^{2}$ )

Hence, other factor of the required epression is $x^{2}$ - y

The expression would be $((x^{2})^{3} - y^{3}) = (x^{6} - y^{3})$

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Similar questions

If ${(x}^{4} + x^{2} y + y^{2})$ is one of the factors of an expression which is the difference of two cubes, then the other factor is ___.

If ( $x^{4} + x^{2} y + y^{2})$ is one of the factors of an expression which is the difference of two cubes, then the other factor is $x^{2} - y$ .

If one of the two factors of an expression which is the difference of two cubes is ( $x^{4}$ + $x^{2}$ y + $y^{2}$ ), then the other factor would be -

I . x^{2} - 463 = 321

I I . y^{2} - 421 = 308

Q. Let

x

be the mean of

x_{1}, x_{2}, . . ., x_{n}

and

y

be the mean of

y_{1}, y_{2}, . . ., y_{n}

.
If

z

is the mean of

x_{1}, x_{2}, . . ., x_{n}, y_{1}, y_{2}, . . ., y_{n},

then

z = ?

(a)

(x + y)

(b)

\frac{1}{2} (x + y)

(c)

\frac{1}{n} (x + y)

(d)

\frac{1}{2 n} (x + y)

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If one of the two factors of an expression which is the difference of two cubes is (x4 + x2y + y2), then the other factor would be -

The correct option is D x2 – y x4 + x2y + y2 = (x2)2 + x2y + (y)2 This is in the form of a2+ ab + b2 where a = x2, b = y The other factor is a – b as a3 – b3 = (a - b)( a2+ ab + b2) Hence, other factor of the required epression is x2 - y The expression would be ((x2)3−y3)=(x6−y3)

If one of the two factors of an expression which is the difference of two cubes is ( $x^{4}$ + $x^{2}$ y + $y^{2}$ ), then the other factor would be -