Question

answer question with photos​

Answer 1

Step-by-step explanation:

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Answer 2

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

$\boxed{\bf{(x - y)(x + y)( {x}^{2} + {y}^{2})( {x}^{4} + {y}^{4}) = {x}^{8} - {y}^{8} }}$

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

$(x - y)(x + y)( {x}^{2} + {y}^{2})( {x}^{4} + {y}^{4})$

$= ((x - y)(x + y))( {x}^{2} + {y}^{2})( {x}^{4} + {y}^{4})$

We know that, (a - b)(a + b) = a² - b²

Here a = x, b = y

By substituting the values in the above identity we have,

$= ( {x}^{2} - {y}^{2})( {x}^{2} + {y}^{2})( {x}^{4} + {y}^{4})$

We know that, (a - b)(a + b) = a² - b²

Here a = x² , b = y²

By substituting the values in the above identity we have,

$= ({( {x}^{2})}^{2} - {( {y}^{2}) }^{2})( {x}^{4} + {y}^{4})$

$= ( {x}^{2 \times 2} - {y}^{2 \times 2} )( {x}^{4} + {y}^{4})$

$= ( {x}^{4} - {y}^{4})( {x}^{4} + {y}^{4})$

$\tt{since \: {( {a}^{m}) }^{n} = {a}^{mn} }$

We know that, (a - b)(a + b) = a² - b²

Here $\sf{a = {x}^{4} \: and \: b = {y}^{4}}$

By substituting the values in the above identity we have,

$= {( {x}^{4})}^{2} - {( {y}^{4}) }^{2}$

$\tt{since \: {( {a}^{m}) }^{n} = {a}^{mn} }$

$= {x}^{4 \times 2} - {y}^{4 \times 2}$

$= {x}^{8} - {y}^{8}$

$\boxed{\bf{(x - y)(x + y)( {x}^{2} + {y}^{2})( {x}^{4} + {y}^{4}) = {x}^{8} - {y}^{8} }}$

$\mathfrak{\large{\underline{\underline{Identity\:Used:-}}}}$

(a - b)(a + b) = a² - b²

$\mathfrak{\large{\underline{\underline{Extra\:Information:-}}}}$

[1] (x + y)² = x² + y² + 2xy

[2] (x - y)² = x² + y² - 2xy

[3] (x + y)(x - y) = x² - y²

[4] (x + a)(x + b) = x² + (a + b)x + ab

$\bf{\large{\underline{\underline{Note:-}}}}$

(x + y)(x - y) = x² - y² or (x - y)(x -

+ y) = x² - y² both are same.