• # question_answer Factorise: (a) ${{a}^{4}}-{{b}^{4}}$                            (b) ${{p}^{4}}-81$ (c) ${{x}^{4}}-{{(y+2)}^{4}}$                  (d) ${{x}^{4}}-{{(x-z)}^{4}}$ (e) ${{a}^{4}}-2{{a}^{2}}{{b}^{2}}+{{b}^{4}}$

 (a) Using   ${{a}^{2}}-{{b}^{2}}=(a-b)(a+b)$ ${{a}^{4}}-{{b}^{4}}={{({{a}^{2}})}^{2}}-{{({{b}^{2}})}^{2}}$ $=({{a}^{2}}+{{b}^{2}})({{a}^{2}}-{{b}^{2}})$ $=({{a}^{2}}+{{b}^{2}})(a+b)(a-b)$ (b)   ${{p}^{4}}-81={{\left( {{p}^{2}} \right)}^{2}}-{{\left( 9 \right)}^{2}}$ $=({{p}^{2}}+9)\left( {{p}^{2}}-9 \right)$ $[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)]~$ $=\left( {{p}^{2}}+9 \right)\left( p-3 \right)\left( p+3 \right)~~$ (c)   ${{x}^{4}}-{{(y+2)}^{4}}={{({{x}^{2}})}^{2}}-{{[{{(y+2)}^{2}}]}^{2}}$ $=[({{x}^{2}})+{{(y+2)}^{2}}]\,\,[({{x}^{2}})-{{(y+2)}^{2}}]$ $=[{{(x)}^{2}}+{{(y+2)}^{2}}][(x-y-z)(x+y+2)]$ (d)   ${{x}^{4}}-{{(x-z)}^{4}}={{({{x}^{2}})}^{2}}-{{[{{(x-z)}^{2}}]}^{2}}$ $=[{{x}^{2}}-{{(x-z)}^{2}}]\,[{{x}^{2}}+{{(x-z)}^{2}}]$ $=\left[ \left( x-x+z \right)\left( x+x-\text{ }z \right) \right]\text{ }\left[ ({{x}^{2}}+{{(x-z)}^{2}} \right]~$ $=\text{ }z\left( 2x-z \right)\text{ }\left[ {{x}^{2}}+{{\left( x \right)}^{2}}+{{\left( z \right)}^{2}}-2xz \right]$ $=z(2x-z)[2{{x}^{2}}-2xz+{{z}^{2}}]$ (e)    ${{a}^{4}}-2{{a}^{2}}{{\text{b}}^{2}}+{{b}^{4}}={{\left( {{a}^{2}} \right)}^{2}}+{{\left( {{b}^{2}} \right)}^{2}}-2\left( {{a}^{2}} \right)\left( {{b}^{2}} \right)$ $={{({{a}^{2}}-{{b}^{2}})}^{2}}$ $=[({{a}^{2}}-{{b}^{2}})({{a}^{2}}+{{b}^{2}})]$ $=[(a-b)(a+b)({{a}^{2}}+{{b}^{2}})]$