A few questions about multiplication formulas

1 (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac

3(a^2+b^2+c^2)-(a+b+c)^2

a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)

a-b)^2+(a-c)^2+(b-c)^2>=0

Therefore, (a+b+c) 2<=3a 2+3b 2+3c 2

2 (a^3+b^3)=(a+b)(a^2+b^2-ab)=(a+b)[(a+b)^2-3ab]=1*(1-3ab)=1-3ab

Therefore a 3 + b 3 + 3 ab = 1 - 3 ab + 3 ab = 1

3 (a^3+b^3)=(a+b)(a^2+b^2-ab)

Therefore a 2 + b 2-ab = 10

a+b)^2-3ab=10

3ab=15 ab=5 substituting a 2 + b 2-ab = 10 to get a 2 + b 2 = 15

4 (a+1/a)^2=25

a^2+2*a*1/a+(1/a)^2=25

get a 2 + (1 a ) 2 = 23

a^3+(1/a)^3=(a+1/a)[a^2-a*1/a+(1/a)^2]=5*(23-1)=110

5 Let these two numbers be n, n+1

Then (n+1) 2-n 2=2n+1 is an odd number.

1:(a+b+c)^2-3(a^2+b^2+c^2)=-2(a^2+b^2+c^2)+2(ab+bc+ac)=-(2a^2+2b^2+2c^2-2ab-2bc-2ac)

a-b) 2-(b-c) 2-(a-c) 2<=0, so (a+b+c) 2<=3(a 2+b 2+c 2), ab=5

a^2+b^2=(a+b)^2-2ab=25-10=154.(a+1/a)^2=25

a^2+2*a*1/a+(1/a)^2=25a^2+(1/a)^2=23

a^3+(1/a)^3=(a+1/a)[a^2-a*1/a+(1/a)^2]=5*(23-1)=110

5.Let the smaller number x

x+1) 2-x 2=(2x+1)(x+1-x)=2x+1, is an odd number.

a+b+c)^2 - 3(a^2+b^2+c^2)-2a^2-2b^2-2c^2+2ab+2bc+2ac-(a-b)^2-(b-c)^2-(a-c)^2≤0(a+b+c)^2 ≤ 3(a^2+b^2+c^2)2.∵a+b=1

a^3+b^3+3ab=(a+1)(a^2+b^2-ab)+3ab(a+b)^2

3.If a+b=5 a 3+b 3=50 find the value of a 2+b 2 a 3 + b 3 = 50

5(a^2+b^2-ab)=50

a^2+b^2-ab=10

a^2+b^2+2ab=25

a^2+b^2=15

4.∵a+1/a=5

a^2+1/a^2+2=25

a^2+1/a^2=23

1/a^3+a^3

1/a+a)(a^2+1/a^2-1)

Then (a+1) 2-a 2

1×(2a+1)

2a+1a is an integer.

2a+1 is an odd number.

1) (a+b+c)^2-3(a^2+b^2+c^2)=2ab+2ac+2bc-2a^2-2b^2-2c^2=-(a-b)^2-(a-c)^2-(b-c)^2<0

2) a^3+b^3+3ab=(a+b)(a^2-ab+b^2)+3ab=a^2-ab+b^2+3ab=(a+b)^2=1

3) a^3+b^3=(a+b)(a^2-ab+b^2)=5(a^2-ab+b^2)=50

a^2-ab+b^2=10---1)

While: (a+b) 2=25

a^2+2ab+b^2=25---2)

From (1), (2), get:

3(a^2+b^2)=2*10+25=45

a^2+b^2=15

4) a^2+a^(1/2)=(a+(1/a))^2-2=5^2-2=23

1/a^3)+a^3=(a+(1/a))(a^2+(1/a^2)-1)=5((a+(1/a))^2-3)=5(5^2-3)=110

5) (n+1) 2-n 2=2n+1, is an odd number.

1, the latter is greater and equal to the former. Because 3(a 2+b 2+c 2)-(a+b+c) 2=(a-b) 2+(b-c) 2+(a-c) 2>=0

2. The answer is.

3,A 3+b 3=(a+b)(a 2+b 2-ab) =50,a 2+b 2-ab=10=(a+b) 2-3ab, AB=5, so a 2+b 2=15

4, a 2 + a square fraction = (a + 1 a) 2-2 = 23a 3 + ( a) 3 = (a + 1 a) (a 2 + 1 a 2-1) = 5 * 22 = 110

5,(n+1) 2-n 2=2n+1, is an odd number.

Hello! Is it 2, if it is: you can bring x=y+z=2 into x 3+2y 3+2z 3+6xyz.

2^3+2*(2-z)^3+2z^3+6*2*(2-z)z=4*3+6*(4-4z+z^)+6z^+12*(2z-z^)=12+24-24z+6z^+6z^+24z-12z^=36

1.(x+6) squared - (6-x) squared = (x+6+6-x)*(x+6-6+x)=12*2x=24x

2, .x+3y)(x-3y)} squared = (x 2-9y 2) 2 = x 4-18x 2y 2+81y 4

3,.（x+2y-5）（x-2y+5）=[x+(2y-5)][x-(2y-5)]=x^2-(2y-5)^2=x^2-4y^2+20y-25

a +b -a -2ab-b, if nothing else, the answer should be this. Maybe the teacher copied the wrong answer, this is a very simple question.