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Let's factorize each expression in Question 3.

### Solutions:

(a) \( x^4 - 1 \)

This is a difference of squares:
\[
x^4 - 1 = (x^2 + 1)(x^2 - 1)
\]
Now, \( x^2 - 1 \) is also a difference of squares:
\[
x^4 - 1 = (x^2 + 1)(x + 1)(x - 1)
\]

---

(b) \( y^4 - 16 \)

This is also a difference of squares:
\[
y^4 - 16 = (y^2 + 4)(y^2 - 4)
\]
Further factoring \( y^2 - 4 \) as a difference of squares:
\[
y^4 - 16 = (y^2 + 4)(y + 2)(y - 2)
\]

---

(e) \( h^2 - p^4 \)

This is a difference of squares:
\[
h^2 - p^4 = (h + p^2)(h - p^2)
\]

---

(f) \( 16x^4 - 49 \)

This is a difference of squares:
\[
16x^4 - 49 = (4x^2 + 7)(4x^2 - 7)
\]

---

(i) \( 81p^4 - x^6 \)

Rewrite this as a difference of squares:
\[
81p^4 - x^6 = (9p^2 + x^3)(9p^2 - x^3)
\]

---

(j) \( 144x^8 - 1 \)

This is a difference of squares:
\[
144x^8 - 1 = (12x^4 + 1)(12x^4 - 1)
\]
Further factoring \( 12x^4 - 1 \) as a difference of squares:
\[
144x^8 - 1 = (12x^4 + 1)(\sqrt{12}x^

Factorize each of the following expressions: (a) x^4 - 1, (b) y^4 - 16, (e) h^2 - p^4, (f) 16x^4 - 49, (i) 81p^4 - x^6, (j) 144x^8 - 1.

Explore step-by-step solutions for factorizing expressions using the difference of squares method. Learn to factorize terms like x^4 - 1, y^4 - 16, and others, ideal for high school students in Grades 9-11.

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