Question

Factorise
(i) ${a^4} - {b^4}$
(ii) ${p^4} - 81$
(iii) ${x^4} - {(y + z)^4}$
(iv) ${x^4} - {(x - z)^4}$
(v) ${a^4} - 2{a^2}{b^2} + {b^4}$

Answer 1

Hint:
The given polynomials are of degree four. We can factorise them using the identities in Algebra. Repeated applications may be needed for simplification.

Useful formula:
For any $a,b$ we have, ${a^2} - {b^2} = (a - b)(a + b)$
${(a + b)^2} = {a^2} + 2ab + {b^2}$
${(a - b)^2} = {a^2} - 2ab + {b^2}$

Complete step by step solution:
We have to factorise the given equations.
(i) ${a^4} - {b^4}$
We can write it as ${({a^2})^2} - {({b^2})^2}$.
For any $a,b$ we have, ${a^2} - {b^2} = (a - b)(a + b)$
So we have, ${({a^2})^2} - {({b^2})^2} = ({a^2} - {b^2})({a^2} + {b^2})$
We can again factorise using the same result.
This gives,
${({a^2})^2} - {({b^2})^2} = (a - b)(a + b)({a^2} + {b^2})$
So we have,
${a^4} - {b^4} = (a - b)(a + b)({a^2} + {b^2})$

(ii) ${p^4} - 81$
This can be written in the form ${({p^2})^2} - {9^2}$
For any $a,b$ we have, ${a^2} - {b^2} = (a - b)(a + b)$
This gives,
${({p^2})^2} - {9^2} = ({p^2} - 9)({p^2} + 9)$
We can again factorise using the same result for ${p^2} - 9$.
We have ${p^2} - 9 = {p^2} - {3^2}$
This gives,
${({p^2})^2} - {9^2} = (p - 3)(p + 3)({p^2} + 9)$
So we have,
${p^4} - 81 = (p - 3)(p + 3)({p^2} + 9)$

(iii) ${x^4} - {(y + z)^4}$
This expression can be rewritten as ${({x^2})^2} - {({(y + z)^2})^2}$
For any $a,b$ we have, ${a^2} - {b^2} = (a - b)(a + b)$
This gives,
${({x^2})^2} - {({(y + z)^2})^2} = ({x^2} - {(y + z)^2})({x^2} + {(y + z)^2})$
Again we can use the same formula for ${x^2} - {(y + z)^2}$.
We have, ${x^2} - {(y + z)^2} = (x - (y + z))(x + (y + z))$
$ \Rightarrow {x^2} - {(y + z)^2} = (x - y - z)(x + y + z)$
This gives,
${({x^2})^2} - {({(y + z)^2})^2} = (x - y - z)(x + y + z)({x^2} + {(y + z)^2})$
We have, ${(a + b)^2} = {a^2} + 2ab + {b^2}$.
We can use this result for ${(y + z)^2}$.
${({x^2})^2} - {({(y + z)^2})^2} = (x - y - z)(x + y + z)({x^2} + {y^2} + 2yz + {z^2})$
So we have,
${x^4} - {(y + z)^4} = (x - y - z)(x + y + z)({x^2} + {y^2} + 2yz + {z^2})$

(iv) ${x^4} - {(x - z)^4}$
We can write this as ${({x^2})^2} - {({(x - z)^2})^2}$
For any $a,b$ we have, ${a^2} - {b^2} = (a - b)(a + b)$
This gives,
${({x^2})^2} - {({(x - z)^2})^2} = ({x^2} - {(x - z)^2})({x^2} + {(x - z)^2})$
Again we can use the same formula for ${x^2} - {(x - z)^2}$.
${x^2} - {(x - z)^2} = (x - (x - z))(x + (x - z))$
Simplifying we get,
${x^2} - {(x - z)^2} = (x - x + z)(x + x - z)$
$ \Rightarrow {x^2} - {(x - z)^2} = z(2x - z)$
This gives,
${({x^2})^2} - {({(x - z)^2})^2} = z(2x - z)({x^2} + {(x - z)^2})$
We have, ${(a - b)^2} = {a^2} - 2ab + {b^2}$.
We can use this result for ${(x - z)^2}$.
$ \Rightarrow {({x^2})^2} - {({(x - z)^2})^2} = z(2x - z)({x^2} + ({x^2} - 2xz + {z^2}))$
Simplifying we get,
$ \Rightarrow {({x^2})^2} - {({(x - z)^2})^2} = z(2x - z)(2{x^2} - 2xz + {z^2})$
So we have,
${x^4} - {(x - z)^4} = z(2x - z)(2{x^2} - 2xz + {z^2})$

(v) ${a^4} - 2{a^2}{b^2} + {b^4}$
We can write it as ${({a^2})^2} - 2{a^2}{b^2} + {({b^2})^2}$
${a^4} - 2{a^2}{b^2} + {b^4} = {({a^2})^2} - 2{a^2}{b^2} + {({b^2})^2}$
We have, ${(a + b)^2} = {a^2} + 2ab + {b^2}$.
Comparing these two we have,
${a^4} - 2{a^2}{b^2} + {b^4} = {({a^2} - {b^2})^2}$
For any $a,b$ we have, ${a^2} - {b^2} = (a - b)(a + b)$
So we have,
${a^4} - 2{a^2}{b^2} + {b^4} = {[(a - b)(a + b)]^2}$

Note:
When an equation of degree two or more is given, we have to factorise maximum to linear factors. The main identities used are the expansions of ${(a + b)^2},{(a - b)^2}$ and ${a^2} - {b^2}$. In certain questions repeated application of the same result may be needed.