Corresponde al cubo del primer término, más (o menos) el triple del cuadrado del primer término multiplicado por el segundo, más el triple del primer término multiplicado por el cuadrado del segundo y más (o menos) el cubo del segundo, así:
Ejercicios: CUBO DEL BINOMIO
1. (a + 3)³= a³ + 3(a)²(3) + 3(a)(3)² + (3)³
= 27 + 9a² + 27a + 27
- 2. (p – q)³ = p³ – 3(p)²(q) + 3(p)(q)² – q³= p³ – 3p²q + 3pq² - q³
3. (x + 2)³ = x³ + 3(x)²(2) + 3(x)(2)² + 2³
= x³ + 6x² + 12x + 8
- 4. (a – 3)³ = a³ + 3(a)²(3) + 3(a)(3)²+ (3)³= a³ + 9a² + 27a + 27
- 5. (t + 4)³ = t³ + 3(t)²(4) + 3(t)(4)² + (4)³ == t³ + 3(t)²(4) + 3(t)(4)² + (4)³= t³ + 12t² + 48t + 64
- 6. (2 – a)³ = 2³ – 3(2)²(a) + 3(2)(a)² – a³= 8 – 12a + 6a² - a³
- 7. (2a – b)³ = (2a)³ -3(2a)²(b)+3(2a)(b)²- b³ =8a³ – 3(4a²)b + 6ab² – b³= 8a³ – 12a²b + 6ab² – b³
- 8. (3a - 5b)³ = (3a)³-3(3a)²(5b)+3(3a)(5b)²-(5b)³= 27a³ - 135a²b + 225ab² - 125b³
9. (2x + 3y)³=(2x)³+3(2x)²(3y)+3(2x)(3y)²+(3y)³ = 8x³ + 36 x²y + 54xy² + 27y³
10. (1 – 3y)³ = (1)³ – 3(1)²(3y)+3(1)(3y)²- (3y)³ = 1 – 9y + 27y² - 27y³
11. (2 + 3t)³ = 2³ + 3 (2)²(3t) + 3(2)(3t)² + (3t)³ = 8 + 36t + 54t² +27t³
- 12. (3a –2x)³=(3a)³–3(3a)²(2x)+3(3a)(2x)²-(2x)³ = 27a³ – 54a²x + 36ax² – 8x³
13. (5a – 1)³= (5 a)³–3(5a)²(1)+3(5a)(1)² - (1)³ =125 a³ – 75a² + 15a - 1
14. (3a²-2a)³=(3a²)³–3(3a²)²(2a)+3(3a²)(2a)²-(2a)³
= (27x6) - 3(9x4) ²(2a)+(9a²)(4a²) - (8a³)
= (27x6) - (54x5) +(36a4) - (8a³)
= (27x6) - 3(9x4) ²(2a)+(9a²)(4a²) - (8a³)
= (27x6) - (54x5) +(36a4) - (8a³)
- 15. (t² + t³)³ =(t²)³+ 3(t²)²(t³) + 3(t²)(t³)² + (t³)³= t⁶ + 3t⁴t³ + 3t²t⁶ + t⁹= t⁶ + 3t⁷ + 3t⁸ + t⁹
- 16. ( 1 + x⁴)³= (1)³ + 3(1)²(x⁴) + 3(1)(x⁴)² + (x⁴)³= 1 + 3x⁴ + 3x⁸ + x¹²
- 17. (2t – 3a²)³=(2t)³–3(2t)²(3a²)+3(2t)(3a²)²– (3a²)³ == 8t³ – 36t²a² + 54ta⁴ - 27a⁶
18. (u² +5v)³=(u²)³+3(u²)²(5v)+3(u²)(5v)²+ (5v)³
= u⁶ + 15u⁴5v + 75u²v² - 125v³
- 19. ( ½ - a)³ = (½ )³ - 3(1/2)²(a) +3(1/2)(a)² – a³
= 1/8 – 3/4 a + 3/2 a² – a³
20. (1/2x + 2y)³=(1/2x)³+3(1/2x)²(2y)+ 3(1/2x)(2y)² + (2y)³
= x³/8 + 6x²y/4 – 12y²x/2 + 8y³
= 1/8 x³+ 3/2 x²y – 6 x y²+ 8 y³
21. (2/3a – 1/3b)³ =
=(2/3a)³–3(2/3a)²(1/3b)+3(2/3a)(1/3b)²+(1/3b)³
= 8a³/27 – 12a²b/27 + 6ab²/27 – 1/27b³
= 8/27 a³ – 4/9 a²b + 2/9 ab² – 1/27 b³
22. (5p/2 + 3q/2)³ =
=(5p/2)³+3(5p/2)²(3q/2)+3(5p/2)(3q/2)²+(3q/2)³
= 125p³/8 + 225p²q/8 + 135pq²/8 + 27q³/8
- 23. ( 1 m/10 – 1 n /5)³ =
= (1m/10)³–3(1m/10)²(1n/5)+3(1m/10)(1n/5)² – (1n/5)³
= m³/1000 – 3mn/500 + 3mn²/250 – n³/125
- 23. (a – a/3)³ = a³–3(a)²(a/3)+3(a)(a/3)² – (a/3)³
= a³ – 3a³/3 + 3a³/9 – a³/27
= a³ – a³ + a³/3 – a³/27
= a³/3 – a³/27
= a³ – a³ + a³/3 – a³/27
= a³/3 – a³/27
- 24. (1t/2 + 2t²)³ =
= (1t/2)³ – 3(1t/2)²(2t²) + 3(1t/2)(2t²)² - (2t²)³ =t³/8 – 6t³/4 + 12t⁵/2 – 8t⁶
=t³/8 – 3t³/2 + 6t⁵ - 8t⁶
