1–a(a–b+c)+b(a–b+c)+c(b–a–c)=
= 1–а(а–b+c)+b(a–b+c)–c(a–b+c)=
=1–(a–b+c)(a–b+c)=
=1–(a–b+c)²= (1–a+b–c)(1+a–b+c).
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1–a(a–b+c)+b(a–b+c)+c(b–a–c)=
= 1–а(а–b+c)+b(a–b+c)–c(a–b+c)=
=1–(a–b+c)(a–b+c)=
=1–(a–b+c)²= (1–a+b–c)(1+a–b+c).