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

Byju's Answer

Standard VIII

Mathematics

Multiplication of Any Polynomial

xv Multiply t...

Question

Question 83 (xv)

Multiply the following:
$(a^{2} - b^{2}), (a^{2} + b^{2})$

Open in App

Solution

We have, $(a^{2} - b^{2}) a n d (a^{2} + b^{2})$
$∴ (a^{2} - b^{2}) (a^{2} + b^{2})$ = $a^{2} (a^{2} + b^{2}) - b^{2} (a^{2} + b^{2})$
$= a^{4} + a^{2} b^{2} - b^{2} a^{2} - b^{2} = a^{4} - b^{4}$

Suggest Corrections

Similar questions

Which of the following is correct?
a) $(a - b)^{2} = a^{2} + 2 a b - b^{2}$
b) $(a - b)^{2} = a^{2} - 2 a b + b^{2}$
c) $(a - b)^{2} = a^{2} - b^{2}$
d) $(a + b)^{2} = a^{2} + 2 a b - b^{2}$

Q. Question 3

Which of the following is correct?
a)

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

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

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

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

Q. Question 4
If

s i n θ = \frac{a}{b}

, then

c o s θ

is equal to

(A)

\frac{b}{\sqrt{b^{2} - a^{2}}}

(B)

\frac{b}{a}

(C)

\frac{\sqrt{b^{2} - a^{2}}}{b}

(D)

\frac{a}{\sqrt{b^{2} - a^{2}}}

Q. Question 90 (vii)

Factorise the following, using the identity,

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

a^{2} y^{2} - 2 a b y + b^{2}

\frac{\sqrt{a^{2} - b^{2}} + a}{\sqrt{a^{2} + b^{2}} + b} \div \frac{\sqrt{a^{2} + b^{2}} - b}{a - \sqrt{a^{2} - b^{2}}}

Join BYJU'S Learning Program

Explore more

Multiplication of Any Polynomial

Standard VIII Mathematics

Join BYJU'S Learning Program

Question 83 (xv) Multiply the following: (a2−b2),(a2+b2)

We have, (a2−b2) and (a2+b2) ∴(a2−b2)(a2+b2) = a2(a2+b2)−b2(a2+b2) =a4+a2b2−b2a2−b2=a4−b4

Question 83 (xv)

Multiply the following:
$(a^{2} - b^{2}), (a^{2} + b^{2})$

We have, $(a^{2} - b^{2}) a n d (a^{2} + b^{2})$
$∴ (a^{2} - b^{2}) (a^{2} + b^{2})$ = $a^{2} (a^{2} + b^{2}) - b^{2} (a^{2} + b^{2})$
$= a^{4} + a^{2} b^{2} - b^{2} a^{2} - b^{2} = a^{4} - b^{4}$