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x2 -y2z +y2 -...

Question

$(x^{2} - y^{2}) z + (y^{2} - z^{2}) x$

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Solution

$(x^{2} - y^{2}) z + (y^{2} - z^{2}) x$
Above terms can be written as,
$z x^{2} - z y^{2} + x y^{2} - x z^{2}$
Rearrange the above terms we get,
$z x^{2} - x z^{2} + x y^{2} - z y^{2}$
Take out common in both terms,
$z x (x - z) + y^{2} (x - z)$
$(x - z) (z x + y^{2})$

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Similar questions

Q. Multiply the monomial by the binomial and find the value of each for x = −1, y = 0.25 and z = 0.05:
(i) 15y²(2 − 3x)
(ii) −3x(y² + z²)
(iii) z²(x − y)
(iv) xz(x² + y²)

Q. Find the angle of intersection of the following curves

(i) y² = x and x² = y
(ii) y = x² and x² + y² = 20
(iii) 2y²= x³ and y² = 32x
(iv) x²+ y² − 4x − 1 = 0 and x² + y² − 2y − 9 = 0
(v)

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

and x² + y² = ab
(vi) x² + 4y² = 8 and x² − 2y² = 2
(vii) x² = 27y and y² = 8x
(viii) x² + y² = 2x and y² = x

x^{2} (y - z) + y^{2} (z - x) + z^{2} (x - y)

x, y, z

from the equations

\frac{x^{2} (y + z)}{a^{3}} = \frac{y^{2} (z + x)}{b^{3}} = \frac{z^{2} (x + y)}{c^{3}} = \frac{x y z}{a b c} = 1

Q. If y is the mean proportional between x and z, prove that :

\frac{x^{2} - y^{2} + z^{2}}{x^{- 2} - y^{- 2} + z^{- 2}} = y^{4}

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