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(x±a)(x±b) = x²±(a+b)x+ab.

Simplify:i a+...

Question

Simplify:

(i) (a +b + c)² + (a − b + c)²

(ii) (a +b + c)² − (a − b + c)2

(iii) (a +b + c)²+ (a − b + c)² + (a +b − c)²

(iv) (2x + p − c)²− (2x − p + c)²

(v) (x² + y²− z²) − (x² − y² + z²)²

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Solution

In the given problem, we have to simplify the expressions

(i) Given

By using identity

Hence the equation becomes

Taking 2 as common factor we get

Hence the simplified value of is

(ii) Given

By using identity

Hence the equation becomes

Taking 4 as common factor we get

Hence the simplified value of is.

(iii) Given

By using identity , we have

Taking 3 as a common factor we get

Hence the value ofis

.

(iv) Given

By using identity , we get

By cancelling the opposite terms, we get

Taking as common a factor we get,

Hence the value of is

(v) We have (x² + y²− z²) − (x² − y² + z²)²

Using formula, we get

(x² + y²− z²) − (x² − y² + z²)²

By canceling the opposite terms, we get

Taking as common factor we get

Hence the value of is.

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Similar questions

\in R^{+}

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

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

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

a, b, c

are positive real number then the system of equations

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1, \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1, \frac{- x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

a, b, c > R^{+}

a, b, c

are positive real numbers) then the following system of equations in

x, y, z

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

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

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

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

\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} .

a, b, c

be the positive numbers. The following system of equations in

x . y

z

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

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

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

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(x±a)(x±b) = x²±(a+b)x+ab.

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