MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Expansion of (x^2 - y^2)

Evaluate : i ...

Question

Evaluate :

(i) $(6 - x y) (6 + x y)$

(ii) $(7 x + \frac{2}{3} y) (7 x - \frac{2}{3} y)$

(iii) $(\frac{a}{2 b} + \frac{2 b}{a}) (\frac{a}{2 b} - \frac{2 b}{a})$

(iv) $(3 x - \frac{1}{2 y}) (3 x + \frac{1}{2 y})$

(v) $(2 a + 3) (2 a - 3) (4 a^{2} + 9)$

(vi) $(a + b c) (a - b c) (a^{2} + b^{2} c^{2})$

(vii) $(5 x + 8 y) (3 x + 5 y)$

(viii) $(7 x + 15 y) (5 x - 4 y)$

(ix) $(2 a - 3 b) (3 a + 4 b)$

(x) $(9 a - 7 b) (3 a - b)$

Open in App

Solution

(i) $(6 - x y) (6 + x y) = 6 (6 + x y) - x y (6 + x y) = 36 + 6 x y - 6 x y + (x y)^{2} = 36 - x^{2} y^{2}$

(ii) $(7 x + \frac{2}{3} y) (7 x - \frac{2}{3} y)$

$= 7 x (7 x - \frac{2}{3} y) (7 x - \frac{2}{3} y)$

$= 49 x^{2} - \frac{14}{3} x y + \frac{14}{3} x y - \frac{4}{9} y^{2} = 49 x^{2} - \frac{4}{9} y^{2}$

(iii) $(\frac{a}{2 b} + \frac{2 b}{a}) (\frac{a}{2 b} - \frac{2 b}{a}) = \frac{a}{2 b} (\frac{a}{2 b} - \frac{2 b}{a}) + \frac{2 b}{a} (\frac{a}{2 b} - \frac{2 b}{a})$

$= \frac{a^{2}}{4 b^{2}} - 1 + 1 - \frac{4 b^{2}}{a^{2}} = \frac{a^{2}}{4 b^{2}} - \frac{4 b^{2}}{a^{2}}$

(iv) $(3 x - \frac{1}{2 y}) (3 x + \frac{1}{2 y}) = 3 x (3 x + \frac{1}{2 y}) - \frac{1}{2 y} (3 x + \frac{1}{2 y})$

$= 9 x^{2} + \frac{3 x}{2 y} - \frac{3 x}{2 y} - \frac{1}{4 y^{2}} = 9 x^{2} - \frac{1}{4 y^{2}}$

(v) $(2 a + 3) (2 a - 3) (4 a^{2} + 9) = [(2 a)^{2} - (3)^{2}] (4 a^{2} + 9) [(a + b) (a - b) = a^{2} - b^{2}]$

$= (4 a^{2} - 9) (4 a^{2} + 9)$

$= (4 a^{2})^{2} - (9)^{2} [(a + b) (a - b) = a^{2} - b^{2}]$

$= 16 a^{4} - 81$

(vi) $(a + b c) (a - b c) (a^{2} + b^{2} c^{2}) = [(a)^{2} - (b c)^{2}] (a^{2} + b^{2} c^{2}) = [(a)^{2} - (b c)^{2}] (a^{2} + b^{2} c^{2})$

$[(a + b) (a - b) = a^{2} - b^{2}]$

$= (a^{2} - b^{2} c^{2}) (a^{2} + b^{2} c^{2})$

$= (a^{2})^{2} - (b^{2} c^{2})^{2} [∵ (a + b) (c - b) = a^{2} - b^{2}]$

$= a^{4} - b^{4} c^{4}$

(vii) $(5 x + 8 y) (3 x + 5 y) = 5 x (3 x + 5 y) + 8 y (3 x + 5 y)$

$= 15 x^{2} + 25 x y + 24 x y + 40 y^{2}$

$= 15 x^{2} + 49 x y + 40 y^{2}$

(viii) $(7 x + 15 y) (5 x - 4 y) = 7 x (5 x - 4 y) + 15 y (5 x - 4 y)$

$= 35 x^{2} - 28 x y + 75 x y - 60 y^{2}$

$= 35 x^{2} + 47 x y - 60 y^{2}$

(ix) $(2 a - 3 b) (3 a + 4 b) = 2 a (3 a + 4 b) - 3 b (3 a + 4 b)$

$= 6 a^{2} + 8 a b - 9 a b - 12 b^{2}$

$= 6 a^{2} - a b - 12 b^{2}$

(x) $(9 a - 7 b) (3 a - b) = 9 a (3 a - b) - 7 b (3 a - b)$

$= 27 a^{2} - 9 a b - 21 a b + 7 b^{2}$

$= 27 a^{2} - 30 a b + 7 b^{2}$

Suggest Corrections

thumbs-up

3

Similar questions

A = - x^{2} - 5 x + 3

B = 4 x^{2} + 7 x - 1

, then calculate

\frac{3}{2} B - 2 A

?

Q. Question 32

The value of

(a + b)^{2} + (a - b)^{2}

2 a + 2 b

2 a - 2 b

2 a^{2} + 2 b^{2}

2 a^{2} + 2 b^{2}

If a and b are the zeroes of the quadratic polynomial f(x)=3x^2-7x-6, find a polynomial whose zeroes are (i) a^2 and b^2 (ii) 2a+3b and 3a+2b

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (1 + a^{2} + b^{2})^{3}

Q. The value of determinant

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

x^2 - y^2

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Expansion of (x^2 - y^2)

Standard VIII Mathematics

Join BYJU'S Learning Program

CrossIcon