Question

Упростите выражение: (b3-b2)(b3+b2)-(1+b2)(1-b2)(1-b2+b4)
И найдите его числовое значение если b=0,1

Answer 1

$(x^a)^c=x^{a*c}$
(x-y)(x+y)=x²-y²
(x+y)(x²-xy+y²)=x³+y³

$(b^3-b^2)(b^3+b^2)-(1+b^2)(1-b^2)(1-b^2+b^4)=b^8-b^4+b^2-1\ \ \ 1)~~(b^3-b^2)(b^3+b^2)=(b^3)^2-(b^2)^2=b^{3*2}-b^{2*2}=b^6-b^4 \ 2)~~ (1+b^1)(1-b^2+b^4)=1^3-(b^2)^3=1-b^6 \3)~~(1-b^2)(1+b^6)=1-b^2+b^6-b^8 \ 4)~~b^6-b^4-(1-b^2+b^6-b^8)=b^6-b^4-1+b^2-b^6+b^8= \ =b^8-b^4+b^2-1 \ \ b=0.1 \ 0.1^8-0.1^4+0.1^2-1=0,00000001-0,0001+0,01-1= \ =-0,99009999$

Answer 2

$( b^{3} - b^{2} )( b^{3} + b^{2} )-(1+ b^{2} )(1- b^{2} )(1 -b^{2}+b^{4} ) = \ =( b^{6} -b^{4} )-(1- b^{4} )(1 -b^{2}+b^{4})= \ =b^{6} -b^{4}-1+ b^{2} -b^{6} + b^{8} =b^{8}-b^{4}+b^{2}-1= \ =0,00000001-0.0001+0,01-1=-0,99009999$

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