Question

find the gcd of 8(x3-x2+x) & 28(x3+1)​

Answer 1

$Given \:two \: Algebraic \:expressions : \\8(x^{3} - x^{2} + x ) \: and \: [28(x^{3}+1)]$

$We \:have ,$

$1) 8(x^{3} - x^{2} + x ) \\=\blue{ 2 \times 2} \times 2 x\blue{(x^{2} - x + 1 ) }$

$2) 28(x^{3} + 1 ) \\= 2\times 2 \times 7 (x^{3} + 1^{3} ) \\= \blue{2\times 2 }\times 7 (x + 1 )\blue{(x^{2} - x + 1 )}$

$\boxed {\pink {Since, a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2} }}$

$Now , G.C.D \: of \: 8(x^{3} - x^{2} + x )\:and \\28(x^{3} + 1 ) \\\blue { = 2\times 2 \:(x^{2} - x + 1 )}$

$\green { = 4(x^{2} - x + 1 )}$

Therefore.,

$\red{ G.C.D \: of \: 8(x^{3} - x^{2} + x )\:and }\\\red{28(x^{3} + 1 ) }\\\green { = 4(x^{2} - x + 1 )}$

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