Question

$Use \: the \: identity \: (x + a) \: (x + b) \\ = {x}^{2} + (a + b) \: x + ab \: to \: \\ find \: the \: following \: product \: .$
1. ( xyz - 4 ) ( xyz - 2 )

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Answer 1

$Soln \: = {(xyz)}^{2} + - 6xyz \: + \: 8$

Step-by-step explanation:

In this Sum :-

• x = xyz
• a = - 4
• b = - 2

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• Subtitue & Solve :-

${(xyz)}^{2} + \: ( - 4) \: xyz \: + ( - 2) \\ {(xyz)}^{2} + \: ( \: ( - 4 \: ) \: + \: ( - 2) \: ) \: xyz \: + \: ( - 4) \: ( - 2) \\ {(xyz)}^{2} + \: - 6xyz \: + \: 8$

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Answer 2

x→xyz,a→−4,b→2

x→xyz,a→−4,b→2Using (x+a)(x+b)=x

x→xyz,a→−4,b→2Using (x+a)(x+b)=x 2

x→xyz,a→−4,b→2Using (x+a)(x+b)=x 2 +(a+b)x+ab

x→xyz,a→−4,b→2Using (x+a)(x+b)=x 2 +(a+b)x+ab(xyz−4)(xyz−2)=(xyz)

x→xyz,a→−4,b→2Using (x+a)(x+b)=x 2 +(a+b)x+ab(xyz−4)(xyz−2)=(xyz) 2

x→xyz,a→−4,b→2Using (x+a)(x+b)=x 2 +(a+b)x+ab(xyz−4)(xyz−2)=(xyz) 2 +(−4+2)xyz+(−4×2)

x→xyz,a→−4,b→2Using (x+a)(x+b)=x 2 +(a+b)x+ab(xyz−4)(xyz−2)=(xyz) 2 +(−4+2)xyz+(−4×2)=x

x→xyz,a→−4,b→2Using (x+a)(x+b)=x 2 +(a+b)x+ab(xyz−4)(xyz−2)=(xyz) 2 +(−4+2)xyz+(−4×2)=x 2

x→xyz,a→−4,b→2Using (x+a)(x+b)=x 2 +(a+b)x+ab(xyz−4)(xyz−2)=(xyz) 2 +(−4+2)xyz+(−4×2)=x 2 y

x→xyz,a→−4,b→2Using (x+a)(x+b)=x 2 +(a+b)x+ab(xyz−4)(xyz−2)=(xyz) 2 +(−4+2)xyz+(−4×2)=x 2 y 2

x→xyz,a→−4,b→2Using (x+a)(x+b)=x 2 +(a+b)x+ab(xyz−4)(xyz−2)=(xyz) 2 +(−4+2)xyz+(−4×2)=x 2 y 2 z

x→xyz,a→−4,b→2Using (x+a)(x+b)=x 2 +(a+b)x+ab(xyz−4)(xyz−2)=(xyz) 2 +(−4+2)xyz+(−4×2)=x 2 y 2 z 2

x→xyz,a→−4,b→2Using (x+a)(x+b)=x 2 +(a+b)x+ab(xyz−4)(xyz−2)=(xyz) 2 +(−4+2)xyz+(−4×2)=x 2 y 2 z 2 −2xyz−8

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