$a^{8} - 2 a^{4} b^{4} + b^{8} =$

(a2-b2)2

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(a4-b4)2

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(a4+b4)2

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Solution

The correct option is B (a4-b4)2
Using Identity: $(A - B)^{2} = A^{2} - 2 A B + B^{2}$
We can say, $A = a^{4}$ and $B = b^{4}$
$\Rightarrow a^{8} - 2 a^{4} b^{4} + b^{8} = (a^{4} - b^{4})^{2}$

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Similar questions

$I f \frac{s i n^{4} θ}{a} + \frac{c o s^{4} θ}{b} = \frac{1}{a + b^{'}}, p r o v e t h a t \frac{s i n^{8} θ}{a^{8}} + \frac{c o s^{8} θ}{b^{8}} = \frac{1}{a + b^{8}}$

Answer the questions based on the given information. Numbers are written on the Chess Board as given

a_{8} = a_{1} + a_{2} + a_{3} + . . . . . . . + a_{7}

b_{8} = b_{1} + b_{2} + b_{3} + . . . . . . + b_{7}

.................
..................

h_{8} = h_{1} + h_{2} + h_{3} + . . . . . + h_{8} =

What is

a_{8} + b_{8} + . . . . + h_{8} =

Q. If

a + b = 1

and

a^{2} + b^{2} = 2

, then find the value of

a^{8} + b^{8}

Factorise:
$a^{8} - 2 a^{4} b^{4} + b^{8}$

Q. Factorise:

2 a^{4} b^{4} - 3 a^{3} b^{5} + 4 a^{2} b^{5}

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a8−2a4b4+b8=

The correct option is B (a4-b4)2Using Identity: (A−B)2=A2−2AB+B2 We can say, A=a4 and B=b4 ⇒a8−2a4b4+b8=(a4−b4)2

$a^{8} - 2 a^{4} b^{4} + b^{8} =$

The correct option is B (a4-b4)2
Using Identity: $(A - B)^{2} = A^{2} - 2 A B + B^{2}$
We can say, $A = a^{4}$ and $B = b^{4}$
$\Rightarrow a^{8} - 2 a^{4} b^{4} + b^{8} = (a^{4} - b^{4})^{2}$