Show that x 2 - 2 x y - y 2 - a 2 - 2 a b - b 2 - x - y - a - b - x - y - a - b

x^2 + 2 xy + y^2 (a^2 + 2ab + b^2)(x + y a + b) (x + y)^2 (a b)^2x + y (a + b) = (x + y (a )) (x + y + (a b))[x + y (a b)] [ ...

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Byju's Answer

Standard VIII

Mathematics

Division of a Polynomial by a Monomial

Show that x...

Question

Show that $\frac{x^{2} + 2 x y + y^{2} - a^{2} + 2 a b - b^{2}}{(x + y - a + b)} = (x + y + a - b)$

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Solution

$\frac{x^{2} + 2 x y + y^{2} - (a^{2} + 2 a b + b^{2})}{(x + y - a + b)}$
$\frac{(x + y)^{2} - (a - b)^{2}}{x + y - (a + b)} = \frac{(x + y - (a -)) (x + y + (a - b))}{[x + y - (a - b)]}$
$[∵ a^{2} - b^{2} = (a + b) (a - b)]$
$= x + y + a - b$ Hence, proved

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Similar questions

If A = $[\begin{matrix} a & b b & a \end{matrix}]$ and $A^{2}$ = $[\begin{matrix} x & y y & x \end{matrix}]$ , then

Q. If

a + i b = \frac{x + i y}{x - i y}

prove that

a^{2} + b^{2} = 1

and

\frac{b}{a} = \frac{2 x y}{x^{2} - y^{2}}

Show that the tangent of an angle between the lines $\frac{x}{a} + \frac{y}{b} = 1$ and $\frac{x}{a} - \frac{y}{b} = 1$ is $\frac{2 a b}{a^{2} - b^{2}}$

\frac{x}{a} = \frac{y}{b} = \frac{z}{c},

prove that

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

Q. Solve for

x

and

y

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

x + y = 2 a b

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Show that x2+2xy+y2−a2+2ab−b2(x+y−a+b)=(x+y+a−b)

x2+2xy+y2−(a2+2ab+b2)(x+y−a+b)(x+y)2−(a−b)2x+y−(a+b)=(x+y−(a−))(x+y+(a−b))[x+y−(a−b)][∵a2−b2=(a+b)(a−b)]=x+y+a−b Hence, proved

Show that $\frac{x^{2} + 2 x y + y^{2} - a^{2} + 2 a b - b^{2}}{(x + y - a + b)} = (x + y + a - b)$

$\frac{x^{2} + 2 x y + y^{2} - (a^{2} + 2 a b + b^{2})}{(x + y - a + b)}$
$\frac{(x + y)^{2} - (a - b)^{2}}{x + y - (a + b)} = \frac{(x + y - (a -)) (x + y + (a - b))}{[x + y - (a - b)]}$
$[∵ a^{2} - b^{2} = (a + b) (a - b)]$
$= x + y + a - b$ Hence, proved