Suggested: (ab)^2=a^2 b^2 - if g is a group such that (ab)^2=a^2 b^2 - a+b+c=13 a^2+b^2+c^2=69 find ab+bc+ca - a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) - show that (a-b)2 (a2+b2) and (ab)2 are in ap - a+b=7 ab=12 then 1/a^2+1/b^2= - a+b=7 ab=12 then a^2+b^2= - a^2+b^2+c^2-ab-bc-ca - (a+b+c)2=a2+b2+c2+2(ab+bc+ca) - ab(x^2+y^2)-xy(a^2+b^2) - if a^2+b^2+c^2=20 and a+b+c=0 find ab+bc+ca - a^2+b^2/ab+1 - if ab+bc+ca=0 then find the value of 1/a^2-bc+1/bc^2-ca+1/c^2-ab - a^3-b^3=(a-b)(a^2+ab+b^2) - (ab)^2=a^2 Browse related:
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