Algebraic sources up to 6-10:
☞ (a+b)²=a²+2ab+b²
☞ (a+b)²=(a-b)²+4ab
☞ (a-b)²=a²-2ab+b²
☞ (a-b)²=(a+b)²-4ab
☞ a²+b²=(a+b)²-2ab
☞ a²+b²=(a-b)²+2ab
☞ a²-b²=(a+b)(a-b)
☞ 2(a²+b²)=(a+b)²+(a-b)²
☞ 4ab=(a+b)²-(a-b)²
☞ ab={(a+b)/2}²-{(a-b)/2}²
☞ (a+b+c)²=a²+b²+c²+2(ab+bc+ca)
☞ (a+b)³= a³+3a²b+3ab²+b³
☞ (a+b)³=a³+b³+3ab(a+b)
☞ a-b)³= a³-3a²b+3ab²-b³
☞ (a-b)³=a³-b³-3ab(a-b)
☞ a³+b³=(a+b)(a²-ab+b²)
☞ a³+b³=(a+b)³-3ab(a+b)
☞ a³-b³=(a-b)(a²+ab+b²)
☞ a³-b³=(a-b)³+3ab(a-b)
☞ (a2+b2+c2)=(a+b+c)2–2(ab+bc+ca)
☞ 2(ab+bc+ca)=(a+b+c)2–(a2+b2+c2)
☞ (a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)
☞ a3+b3+c3–3abc=(a+b+c)(a2+b2+c2–ab–bc–ca)
☞ a3+b3+c3–3abc=(a+b+c){(a–b)2+(b–c)2+(c–a)2}
☞ (x+a)(x+b)=x2+(a+b)x+ab
☞ (x+a)(x–b)=x2+(a–b)x–ab
☞ (x–a)(x+b)=x2+(b–a)x–ab
☞ (x–a)(x–b)=x2–(a+b)x+ab
☞ (x+p)(x+q)(x+r)=x3+(p+q+r)x2+(pq+qr+rp)x+pqr
wow....
good!
thanks
Congratulations @kawsar7! You have completed the following achievement on the Steem blockchain and have been rewarded with new badge(s) :
Click on the badge to view your Board of Honor.
If you no longer want to receive notifications, reply to this comment with the word
STOPDo not miss the last post from @steemitboard: