# NCERT Solutions Class 8 Mathematics Solutions for Exercise 14.2 in Chapter 14 - Factorisation

Question 21 Exercise 14.2

Q4) Factorise:

(i) a^4-b^4

(ii) p^4-81

(iii) x^4-\left(y+z\right)^4

(iv) x^4-\left(x-z\right)^4

(v) a^4-2a^2b^2+b^4

Solution 4:

(i) \left(a^2\right)^2-\left(b^2\right)^2

= \left(a^2+b^2\right)\left(a^2-b^2\right)

(ii) \left(p^2\right)^{2\ }-9^2

\left(p^2+9\right)\left(p^2-9\right)

(iii) \left(x^2\right)^2-\left[\left(y+z\right)^2\right]^2

Let x^2=\ m\ and\ \left(y+z\right)^2=n\ \left(y^2+z^2+2yz=\ n\right)

= m^2-n^2

=(m+n)(m-n)

\left(x^2+y^2+z^2+2yz\right)\left(x^2-y^2-z^2-2yz\right)

\left[x^2+\left(y+z\right)^2\right]\left[x^2-\left(y+z\right)^2\right]

x^2+\left(y+z\right)^2\left[\left(x+y+z\right)\left(x-y-z\right)\right]

(iv) \left(x^2\right)^2-\left[\left(x-z\right)^2\right]^2

Let x^2=m\ and \left(x-z\right)^2\ =n\ ,\ n\ =x^2+z^2-2xz

Therefore, m^2-n^2

= (m+n) (m-n)

=\left(x^2+x^2+z^2-2xz\right)\left(x^2-x^2-z^2+2zx\right)

= \left[x^2+\left(x-z\right)^2\right]\left[x^2-\left(x-z\right)^2\right] ....... equation 1

Applying a^2-b^2=\ \left(a-b\right)\left(a+b\right) to 1

\left[x^2+\left(x-z\right)^2\right]\left[\left(x+x-z\right)\left(x-x+z\right)\right]

\left[x^2+\left(x-z\right)^2\right]\left[\left(2x-z\right)\right]

z\left(2x-z\right)\left[x^2+\left(x-z\right)^2\right]

(v) \left(a^2\right)^2-2a^2b^2+\left(b^2\right)^2

\left(a^2-b^2\right)^2

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