*Notes to a video lecture on http://www.unizor.com*

__Algebra+ 04__

*Problem A*

Prove that

if

**X+Y+Z=1**then

*.*

**X²+Y²+Z² ≥ 1/3***Hint A1*

*X² + Y² ≥ 2·X·Y*.

*Hint A2*

*X²+Y²+Z² =*

= (X−a)²+(Y−a)²+(Z−a)²+

+2·a·(X+Y+Z)−3a² ≥

≥ 2a −3a²

= (X−a)²+(Y−a)²+(Z−a)²+

+2·a·(X+Y+Z)−3a² ≥

≥ 2a −3a²

Quadratic polynomial

*2a−3a²*has roots

*a=0*and

*a=2/3*and maximum at

*a=1/3*with value

*2·(1/3)−3·(1/3)²=1/3*.

*Problem B*

Solve an equation

**X**^{8}+ X^{4}+ 1 = 0*Hint B1*

Substitute

*y=X*.

^{4}*Hint B2*

Represent the left part as a product of

*4*polynomials of a second degree by adding and subtracting

*X*.

^{4}*Answer B*

**X**_{1,2,3,4}= ±1/2 ± i·√3/2

**X**_{5,6,7,8}= ±√3/2 ± i/2*Problem C*

Find a number from

*10*to

*99*knowing that its square equals to sum of its digits in cube.

*Hint C*

Actually, we have to solve the equation

*(10·X+Y)² = (X+Y)³*

for natural

*1 ≤ X ≤ 9*and

*0 ≤ Y≤ 9*.

*Answer C*

The only solution is number

*.*

**27**Check:

*27² = 729*

*(2+7)³ = 729*

## No comments:

Post a Comment