Today our topic of discussion is Formulae of Cubes.
Formulae of Cubes
Formulae of Cubes
Formula 6. (a + b)³ = a ³ + 3a ² * b + 3a * b ² + b ³ = a³ + b³+ 3ab(a + b)
Proof: (a + b)³ = (a + b) * (a + b) ²= (a + b)(a ² + 2ab + b ²)
= a(a² + 2ab + b²) + b(a ² + 2ab + b²)
= a³+ 2a² * b + a * b² + a² * b + 2a * b² + b³
= a ³+ 3a² * b + 3a * b² + b ³
= a ³ + b ³ + 3ab(a + b)
Corollary 9. a ³ + b ³ = (a + b) ³ – 3ab(a + b)
Formula 7. (a – b)³ = a ^ 3 – 3a ^ 2 * b + 3a * b ^ 2 – b ³ = a ³- b³ – 3ab(a – b)
Proof: (a – b) ³ = (a – b) * (a – b)²
= (a – b)(a ² – 2ab + b ²)
= a(a ²- 2ab + b²) – b(a ² – 2ab + b²)
= a ³ – 2a ² * b + a * b² – a ² * b + 2a * b ² – b ³
= a ³ – 3a² * b + 3a * b² – b ³
= a ³ – b ³ – 3ab(a – b)
Note: Substituting -b instead of b in Formula 6 we get Formula 7:
\{a + (- b)\} ³ = a ³ + (- b) ^ 3 + 3a(- b)\{a + (- b)\} That is, (a – b)³ = a ³ – b ³ – 3ab(a – b)
Corollary 10. a ³ – b ³ = (a – b)³ + 3ab(a – b)
Formula 8. a ³ + b ³ = (a + b)(a ² – ab + b²)
Proof: a ³+ b ³ = (a + b)³ – 3ab(a + b)
= (a + b)\{(a + b)²- 3ab\} = (a + b)(a ² + 2ab + b² – 3ab)
= (a + b)(a² – ab + b²)
Formula 9. a ³ – b ³= (a – b)(a ² + ab + b²)
Proof: a ³ – b³ = (a – b)³ + 3ab(a – b)
= (a – b)\{(a – b)² + 3ab\}
= (a – b)(a² – 2ab + b² + 3ab) = (a – b)(a ² + ab + b²)
Example 12. Find the cube of 2x + 3 .
Solution: (2x + 3y)³
= (2x)³ + 3 * (2x) ²* 3y + 3 * 2x * (3y)² + (3y)³
= 8x ³ + 3 * 4x ^ 2 * 3y + 3 * 2x * 9y ^ 2 + 27y³
= 8x³ + 36x²* y + 54x * y² + 27y³
Example 13. Find the cube of 2x – y
Solution: (2x – y) ³
= (2x)³ – 3 * (2x) ^ 2 * y + 3 * 2x * y ^ 2 – y³
= 8x³- 3 * 4x ²* y + 3 * 2x * y ² – y ³
= 8x³ – 12x ² * y + 6x * y ² – y³
Example 14. If x = 37 what is the value of 8x³ + 72x ² + 216x +216?
Solution: 8x ³ + 72x ²+ 216x + 216
= (2x)³ + 3 * (2x) ²* 6 + 3 * 2x * (6) ² + (6)³
= (2x + 6) ³ = (2 * 37 + 6) ³ [substituting the values]
= (74 + 6) ³ = (80) ³ = 512000
Example 15. If x – y = 8 and xy = 5 what is the value of x ³ – y ³ + 8 * (x + y) ²
Solution: x ³ – y ³ + 8 * (x + y) ²
= (x – y)³ + 3xy(x – y) + 8\{(x – y) ² + 4xy\}
= (8)³ + 3 * 5 * 8 + 8(8 ^ 2 + 4 * 5) [substituting the values]
= 8³+ 15 * 8 + 8(8 ^ 2 + 4 * 5)
= 8³ + 15 * 8 + 8 * 84
= 8(8 ² + 15 + 84) = 8(64 + 15 + 84)
= 8 * 163
= 1304
Example 16. If a = √3 + √2 , prove that, a³ + 1/(a³) = 18sqrt(3)
Solution: Given that, a = √3 + √2
1/a = 1/√3 + √2
= √3 – √2/√3 + √2√3 – √2 multiplying numerator and denominator by √3 – √2 ]
= √3 – √2/(√3)² – (√2)²) = √3- √2/(3 – 2) = √3 – √2
a + 1/a =√3 + √2 + √3 – √2 = √3 + √2 + √3 – √2 = 2√3
Now t, a³ + 1/(a³) = (a + 1/a)³ – 3a * 1/a * (a + 1/a)
= 2√3 = (2/3)³ — 3(2√3) | : a + 1 = 2 – a
= 2³ * (√3) ³ – 3 * 2√(3) = 8 * 3√3 – 6√(3)
= 24√(3) – 6√3) = 18√(3) Proved)
Example 17. If x + y = 5 , xy = 6 and x > y
1) Find the value of 2(x² + y²) .
2) Find the value of x ³ – y ³ – 3(x² + y ²)
3) Find the value of x5 + y5
Solution:
1) We know, 2(x ² + y ²) = 2\{(x + y)² – 2xy\} = 2(5² – 2 * 6) = 2 * 13 = 26
2(x² + y²) = 26
2) Given that, x + y = 5 and xy = 6 x > y (By the given condition negative value is not acceptable)
x – y = √(x + y) ² – 4xy)
= √(5² – 4 * 6) = √(25 – 24)
= √(1) = 1 x³ – y³ – 3(x² + y ²) = 1³ + 3 * 6 * 1 – 3 2 *
=1+18-39 26 or, (- 1)³ + 3 * 6(- 1) – 3/2 * 26
= (x – y)³ + 3xy(x – y) – 3/2 * 2(x² + y²)
=-20
x ³- y³- 3(x² + y ²) = – 20
3) x + y = 5 and x – y = 1 adding, 2x = 6 subtracting, 2y = 4 y = 4/2 = 2
x = 6/2 = 3
.’ x5 + y5 = 35 + 25 = 243 + 32 = 275
Work:
1) If x = – 2 , what is the value of 27x³- 54x² + 36x – 8 ? 2) If a + b = 5 and ab = 6 find the value of a³ + b³ + 4 * (a – b)²
3) If x = √(5) + √3) find the value of x³ + 8/(x ³)
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