Simple calculation of 49 times 51 plus 451 times 26 plus 25 times 451

Simple calculation of 49 times 51 plus 451 times 26 plus 25 times 451

49*51+(451*26+25*451)=49*51+51*451=51*500=25500

49 * 99-51 simple calculation

49×99-51=49×99+49-100=49×100-100=48×100=4800

How to calculate 9999 times 27 minus 3333 times 51 minus 6666 times 15

9999 times 27 minus 3333 times 51 minus 6666 times 15
=3333×(3×27-51-2×15)
=3333×(81-51-30)
=3333×0
=0

It is known that the quadratic equation AX ^ + BX + C = 0 has two real roots, and the sum of the cubes of the real roots is S1, the sum of the squares of the two roots is S2, and the sum of the two roots is S3
Proving AS1 + BS2 + CS3 = 0

Let these two real roots be X1 and X2, then there is
x1 x2=-b/(2a)
x1x2=c/a
therefore
S1=x1^3 x2^3
=(x1 x2)(x1^2-x1x2 x2^2)
=(x1 x2)(x1^2 2x1x2-3x1x2 x2^2)
=(x1 x2)[(x1 x2)^2-3x1x2]
=[-b/(2a)]{[-b/(2a)]^2-3c/a}
S2=x1^2 x2^2
=x1^2 x2^2 2x1x2-2x1x2
=(x1 x2)^2-2x1x2
=[-b/(2a)]^2-2c/a
S3=x1 x2
=-b/(2a)
that
aS1 bS2 cS3
…………
Replace S1, S2, S3 with the expressions represented by a, B, C into the above expressions

If a

The simplest way to put halo into specific numbers is a = 2, C = 4, B = 3, S1 = 2 / 3, S2 = 1 / 2, S3 = 2 / 3
So S1 = S3 > S2