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Without actually calculating the cube fi...

Without actually calculating the cube find the value of the following :
`(i) (-9)^(3)+(4)^(3)+(5)^(3) " " (ii) (-18)^(3)+(9)^(3)+(9)^(3) " " (iii) (16)^(3)+(1)^(3)+(-17)^(3)`
`(iv) (8)^(3)+(3)^(3)+(-11)^(3)`.

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The correct Answer is:
To solve the given problems without calculating the cubes directly, we will use the identity for the sum of cubes. The identity states: \[ a^3 + b^3 + c^3 = (a + b + c)^3 - 3(a + b + c)(ab + ac + bc) + 3abc \] From this identity, if \(a + b + c = 0\), then: \[ a^3 + b^3 + c^3 = 3abc \] Now, let's apply this to each part of the question. ### (i) \((-9)^3 + (4)^3 + (5)^3\) 1. **Identify \(a\), \(b\), and \(c\)**: - \(a = -9\), \(b = 4\), \(c = 5\) 2. **Calculate \(a + b + c\)**: \[ a + b + c = -9 + 4 + 5 = 0 \] 3. **Since \(a + b + c = 0\)**, we use the identity: \[ (-9)^3 + (4)^3 + (5)^3 = 3(-9)(4)(5) \] 4. **Calculate \(3abc\)**: \[ 3(-9)(4)(5) = 3 \times -9 \times 20 = -540 \] **Final Answer**: \((-9)^3 + (4)^3 + (5)^3 = -540\) ### (ii) \((-18)^3 + (9)^3 + (9)^3\) 1. **Identify \(a\), \(b\), and \(c\)**: - \(a = -18\), \(b = 9\), \(c = 9\) 2. **Calculate \(a + b + c\)**: \[ a + b + c = -18 + 9 + 9 = 0 \] 3. **Since \(a + b + c = 0\)**: \[ (-18)^3 + (9)^3 + (9)^3 = 3(-18)(9)(9) \] 4. **Calculate \(3abc\)**: \[ 3(-18)(9)(9) = 3 \times -18 \times 81 = -4374 \] **Final Answer**: \((-18)^3 + (9)^3 + (9)^3 = -4374\) ### (iii) \((16)^3 + (1)^3 + (-17)^3\) 1. **Identify \(a\), \(b\), and \(c\)**: - \(a = 16\), \(b = 1\), \(c = -17\) 2. **Calculate \(a + b + c\)**: \[ a + b + c = 16 + 1 - 17 = 0 \] 3. **Since \(a + b + c = 0\)**: \[ (16)^3 + (1)^3 + (-17)^3 = 3(16)(1)(-17) \] 4. **Calculate \(3abc\)**: \[ 3(16)(1)(-17) = 3 \times 16 \times -17 = -816 \] **Final Answer**: \((16)^3 + (1)^3 + (-17)^3 = -816\) ### (iv) \((8)^3 + (3)^3 + (-11)^3\) 1. **Identify \(a\), \(b\), and \(c\)**: - \(a = 8\), \(b = 3\), \(c = -11\) 2. **Calculate \(a + b + c\)**: \[ a + b + c = 8 + 3 - 11 = 0 \] 3. **Since \(a + b + c = 0\)**: \[ (8)^3 + (3)^3 + (-11)^3 = 3(8)(3)(-11) \] 4. **Calculate \(3abc\)**: \[ 3(8)(3)(-11) = 3 \times 8 \times -33 = -792 \] **Final Answer**: \((8)^3 + (3)^3 + (-11)^3 = -792\) ### Summary of Answers: 1. \((-9)^3 + (4)^3 + (5)^3 = -540\) 2. \((-18)^3 + (9)^3 + (9)^3 = -4374\) 3. \((16)^3 + (1)^3 + (-17)^3 = -816\) 4. \((8)^3 + (3)^3 + (-11)^3 = -792\)
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NAGEEN PRAKASHAN ENGLISH-POLYNOMIALS-Exercise 2 E
  1. Evaluate without multiplying directly : (i) 33xx27 " " (ii) 103xx9...

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  2. Expand : (i) (3a-5b)^(2) " " (ii) (a+(1)/(a))^(2) " " (iii) (2x-(...

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  3. Expand : (i) (a+b-c)^(2) " " (ii) (a-2b-5c)^(2) " " (iii) (3a-2b...

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  4. Evaluate using formula : (i) (188)^(2) " " (ii) (9.4)^(2) " " (i...

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  5. (i) If a^(2)+b^(2)+c^(2)=20 " and" a+b+c=0, " find " ab+bc+ac. (ii) ...

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  6. Expand : (i) (2x+3y)^(3) " " (ii) (5y-3x)^(3) " " (iii) (2a+3b)...

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  7. Evaluate (2x-3y+5)^(3).

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  8. If a+2b=5, then show that a^(3)+8b^(3)+30ab=125.

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  9. If 2x-3y=10 and xy=16, find the value of 8x^(3)-27y^(3).

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  10. Evaluate : (i) (98)^(3) " " (ii) (598)^(3) " " (iii) (1003)^(3)

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  11. Factorise : 4a^(2)+9b^(2)+16c^(2)+12ab-24bc-16ca

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  12. Verify : (i) x^3+y^3=(x+y)(x^2-x y+y^2) (ii) x^3-y^3=(x-y)...

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  13. Factorise : (i) 9a^(3)-27b^(3) " " (ii) a^(3)-343 " " (iii) a^(...

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  14. Find the product : (i) (x+3)(x^(2)-3x+9) " " (ii) (7+5b)(49-35b+2...

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  15. Factorise : (i) a^(3)+27b^(3)+8c^(3)-18abc " " (ii) 2sqrt(2)a^(3)...

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  16. Find the product : (i) (a+2b+4c)(a^(2)+4b^(2)+16c^(2)-2ab-8bc-4ca) ...

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  17. Factorise : (i) (x-y)^(3)+(y-z)^(3)+(z-x)^(3) (ii) (x-2y)^(3)+(2y-...

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  18. Without actually calculating the cube find the value of the following ...

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  19. Verify that x^3+y^3+z^3-3xyz=1/2(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]

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  20. If x+y+z=0show that x^3+y^3+z^3=3x y z.

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