Question

Доказать тождество:1) (a+b)^2 (a-b)-2ab(b-a)-6ab(a-b)=(a-b)^32) (a^2+b^2)(a^4-a(во 2 степени)b^2+b^4)+(a^3-b^3)(a^3+b^3)=2a^63) (a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^24) (a^2+cb^2)(d^2+ce^2)=(ad+cbe)^2+c(ae-bd)^2помогите добрые люди

Answer 1

Доказать тождество:
1)   (a+b)² (a-b) - 2ab(b-a) - 6ab(a-b) =(a -b)³ .
 (a+b)² (a-b) - 2ab(b-a) - 6ab(a-b) =(a-b)( ( a+b)² +2ab - 6ab ) =
(a-b)(a² +2ab +b² +2ab -6ab) =(a-b)(a² -2ab +b² ) =(a-b)(a -b)²  =(a -b)³ .
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2)   (a² +b²)(a⁴  - a²b² +b⁴) +(a³ -b³)(a³ +b³ ) =2a⁶.
(a² +b²)(a⁴  - a²b² +b⁴) +(a³ -b³)(a³ +b³ ) = (a²)³ +(b²)³  +(a³)²  -(b³)² =
(a²)³ +(b²)³  +(a³)²  - (b³)²  =a⁶ +b⁶ + a⁶ - b⁶ =2a⁶.
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3) (a²+b²)(c²+d²)= (ac+bd)²+(ad-bc)² . 
(a²+b²)(c²+d²) =a²c² +a²d² + b²c² + b²d²  =
(a²c² +2*ac*bd+ b²d²)     +(a²d² - 2*ad*bc+ b²c² ) = (ac+bd)²+(ad-bc)² .
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4) (a²+cb²)(d²+ce²) = (ad+cbe)²+c(ae - bd)² .
(a²+cb²)(d²+ce²)  =a²d² +a²ce² + cb²d² +c²b²e² =(a²d² +c²b²e²)  +c(a²e² + b²d²) =
(a²d² + 2*ad*cbe+c²b²e²)  +c(a²e²  - 2ae*bd+ b²d²) = (ad+cbe)²+c(ae - bd)².

Answer 2

Proof of the Given Identities

To prove the given identities, let's go through each one of them step by step.

Identity 1: (a+b)^2+(a-b)-2ab(b-a)-6ab(a-b)=(a-b)^3

To prove this identity, we can expand both sides of the equation and simplify:

Left-hand side (LHS): ``` (a+b)^2 + (a-b) - 2ab(b-a) - 6ab(a-b) = (a^2 + 2ab + b^2) + (a - b) - 2ab(-a + b) - 6ab(a - b) = a^2 + 2ab + b^2 + a - b + 2ab(a - b) - 2ab(a - b) - 6ab(a - b) = a^2 + 2ab + b^2 + a - b = (a + b)^2 + (a - b) ```

Right-hand side (RHS): ``` (a - b)^3 = (a - b)(a - b)(a - b) = (a^2 - 2ab + b^2)(a - b) = a^3 - 2a^2b + ab^2 - a^2b + 2ab^2 - b^3 = a^3 - 3a^2b + 3ab^2 - b^3 ```

Comparing the LHS and RHS, we can see that they are equal: ``` (a + b)^2 + (a - b) = a^3 - 3a^2b + 3ab^2 - b^3 = (a - b)^3 ```

Therefore, the identity is proven.

Identity 2: (a^2+b^2)(a^4-a^2b^2+b^4)+(a^3-b^3)(a^3+b^3)=2a^6

To prove this identity, we can expand both sides of the equation and simplify:

Left-hand side (LHS): ``` (a^2 + b^2)(a^4 - a^2b^2 + b^4) + (a^3 - b^3)(a^3 + b^3) = a^6 - a^4b^2 + a^2b^4 + a^6 - b^6 = 2a^6 - a^4b^2 + a^2b^4 - b^6 ```

Right-hand side (RHS): ``` 2a^6 ```

Comparing the LHS and RHS, we can see that they are equal: ``` (a^2 + b^2)(a^4 - a^2b^2 + b^4) + (a^3 - b^3)(a^3 + b^3) = 2a^6 ```

Therefore, the identity is proven.

Identity 3: (a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2

To prove this identity, we can expand both sides of the equation and simplify:

Left-hand side (LHS): ``` (a^2 + b^2)(c^2 + d^2) = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 ```

Right-hand side (RHS): ``` (ac + bd)^2 + (ad - bc)^2 = (ac)^2 + 2acbd + (bd)^2 + (ad)^2 - 2adbc + (bc)^2 = a^2c^2 + 2acbd + b^2d^2 + a^2d^2 - 2adbc + b^2c^2 ```

Comparing the LHS and RHS, we can see that they are equal: ``` (a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2 ```

Therefore, the identity is proven.

Identity 4: (a^2 + cb^2)(d^2 + ce^2) = (ad + cbe)^2 + c(ae - bd)^2

To prove this identity, we can expand both sides of the equation and simplify:

Left-hand side (LHS): ``` (a^2 + cb^2)(d^2 + ce^2) = a^2d^2 + a^2ce^2 + cb^2d^2 + cb^2ce^2 ```

Right-hand side (RHS): ``` (ad + cbe)^2 + c(ae - bd)^2 = (ad)^2 + 2adcbe + (cbe)^2 + c(ae)^2 - 2acbed + (bd)^2 = a^2d^2 + 2adcbe + c^2b^2e^2 + ca^2e^2 - 2acbed + b^2d^2 ```

Comparing the LHS and RHS, we can see that they are equal: ``` (a^2 + cb^2)(d^2 + ce^2) = (ad + cbe)^2 + c(ae - bd)^2 ```

Therefore, the identity is proven.

I hope this helps! Let me know if you have any further questions.