factorise”[tex]64a {}^{6} – b {}^{6} [/tex]can you give me this ans in a simple way About the author Allison
Question : Factorise : [tex]\tt 64{a}^{2} – {b}^{2}[/tex] Identity : [tex]\tt {x}^{2} – {y}^{2} [/tex] = (x-y)(x+y) Solution : [tex] \tt 64{a}^{2} – {b}^{2} [/tex] [tex] \tt \implies {(8a)}^{2} – {b}^{2} [/tex] [tex]\tt \implies [/tex] (8a-b)(8a+b) [tex]\tt 64{a}^{2} – {b}^{2}[/tex] = (8a-b)(8a+b) Some other identities : ● [tex] \tt {(a+b)}^{2} = {a}^{2} + 2ab + {b}^{2} [/tex] ● [tex] \tt {(a-b)}^{2} = {a}^{2} – 2ab + {b}^{2} [/tex] ● [tex] \tt {(a+b)}^{3} = {a}^{3} + 3{a}^{2}b + 3a{b}^{2} + {b}^{3} [/tex] ● [tex] \tt {(a-b)}^{3} = {a}^{3} – 3{a}^{2}b + 3a{b}^{2} – {b}^{3} [/tex] Reply
Answer: [tex]\huge \bold {\fbox{\underline \orange{ Answer \: ❦}}}[/tex] [tex]64 {a}^{2} – {b}^{2} = ({4a}^{2})^{3} – ({b}^{2})^{3} \\ \\ = ( {4a}^{2} – {b}^{2})(16 {a}^{4} + 4 {a}^{2} {b}^{2} + {b}^{4} ) \\ \\ = (2a + b)(2a – b) (16 {a}^{4} + 4 {a}^{2} {b}^{2} + {b}^{4} )[/tex] Reply
Question :
Factorise :
[tex]\tt 64{a}^{2} – {b}^{2}[/tex]
Identity :
[tex]\tt {x}^{2} – {y}^{2} [/tex] = (x-y)(x+y)
Solution :
[tex] \tt 64{a}^{2} – {b}^{2} [/tex]
[tex] \tt \implies {(8a)}^{2} – {b}^{2} [/tex]
[tex]\tt \implies [/tex] (8a-b)(8a+b)
[tex]\tt 64{a}^{2} – {b}^{2}[/tex] = (8a-b)(8a+b)
Some other identities :
● [tex] \tt {(a+b)}^{2} = {a}^{2} + 2ab + {b}^{2} [/tex]
● [tex] \tt {(a-b)}^{2} = {a}^{2} – 2ab + {b}^{2} [/tex]
● [tex] \tt {(a+b)}^{3} = {a}^{3} + 3{a}^{2}b + 3a{b}^{2} + {b}^{3} [/tex]
● [tex] \tt {(a-b)}^{3} = {a}^{3} – 3{a}^{2}b + 3a{b}^{2} – {b}^{3} [/tex]
Answer:
[tex]\huge \bold {\fbox{\underline \orange{ Answer \: ❦}}}[/tex]
[tex]64 {a}^{2} – {b}^{2} = ({4a}^{2})^{3} – ({b}^{2})^{3} \\ \\ = ( {4a}^{2} – {b}^{2})(16 {a}^{4} + 4 {a}^{2} {b}^{2} + {b}^{4} ) \\ \\ = (2a + b)(2a – b) (16 {a}^{4} + 4 {a}^{2} {b}^{2} + {b}^{4} )[/tex]