বিসমিল্লহির রহমানির রহিম

ষষ্ঠ শ্রেনী থেকে দশম  বী জগা ণিতিক সূত্রাবলী ☞ (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 – (...