Question

factorise"
$64a {}^{6} - b {}^{6}$
can you give me this ans in a simple way ​

Answer 1

Question :

Factorise :

$\tt 64{a}^{2} - {b}^{2}$

Identity :

$\tt {x}^{2} - {y}^{2}$ = (x-y)(x+y)

Solution :

$\tt 64{a}^{2} - {b}^{2}$

$\tt \implies {(8a)}^{2} - {b}^{2}$

$\tt \implies$ (8a-b)(8a+b)

$\tt 64{a}^{2} - {b}^{2}$ = (8a-b)(8a+b)

Some other identities :

● $\tt {(a+b)}^{2} = {a}^{2} + 2ab + {b}^{2}$

● $\tt {(a-b)}^{2} = {a}^{2} - 2ab + {b}^{2}$

● $\tt {(a+b)}^{3} = {a}^{3} + 3{a}^{2}b + 3a{b}^{2} + {b}^{3}$

● $\tt {(a-b)}^{3} = {a}^{3} - 3{a}^{2}b + 3a{b}^{2} - {b}^{3}$

Answer 2

Answer:

$\huge \bold  {\fbox{\underline \orange{ Answer \: ❦}}}$

$64 {a}^{2} - {b}^{2} = ({4a}^{2})^{3} - ({b}^{2})^{3} \\ \\ = ( {4a}^{2} - {b}^{2})(16 {a}^{4} + 4 {a}^{2} {b}^{2} + {b}^{4} ) \\ \\ = (2a + b)(2a - b) (16 {a}^{4} + 4 {a}^{2} {b}^{2} + {b}^{4} )$