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

Factorise :
`(i) (x-y)^(3)+(y-z)^(3)+(z-x)^(3)`
`(ii) (x-2y)^(3)+(2y-4z)^(3)+(4z-x)^(3)`
`(iv) (3sqrt(2)a-5sqrt(3)b)^(3)+(5sqrt(3)b-7sqrt(5)c)^(3)+(7sqrt(5)c-3sqrt(2)a)^(3)`.

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The correct Answer is:
To factorize the given expressions, we will use the identity that states if \( a + b + c = 0 \), then: \[ a^3 + b^3 + c^3 = 3abc \] Let's solve each part step by step. ### Part (i): Factorize \( (x-y)^3 + (y-z)^3 + (z-x)^3 \) 1. **Identify \( a, b, c \)**: - Let \( a = x - y \) - Let \( b = y - z \) - Let \( c = z - x \) 2. **Check if \( a + b + c = 0 \)**: \[ a + b + c = (x - y) + (y - z) + (z - x) = 0 \] Since the sum is zero, we can use the identity. 3. **Apply the identity**: \[ (x - y)^3 + (y - z)^3 + (z - x)^3 = 3(x - y)(y - z)(z - x) \] ### Final Answer for Part (i): \[ (x - y)^3 + (y - z)^3 + (z - x)^3 = 3(x - y)(y - z)(z - x) \] --- ### Part (ii): Factorize \( (x-2y)^3 + (2y-4z)^3 + (4z-x)^3 \) 1. **Identify \( a, b, c \)**: - Let \( a = x - 2y \) - Let \( b = 2y - 4z \) - Let \( c = 4z - x \) 2. **Check if \( a + b + c = 0 \)**: \[ a + b + c = (x - 2y) + (2y - 4z) + (4z - x) = 0 \] The sum is zero, so we can use the identity. 3. **Apply the identity**: \[ (x - 2y)^3 + (2y - 4z)^3 + (4z - x)^3 = 3(x - 2y)(2y - 4z)(4z - x) \] ### Final Answer for Part (ii): \[ (x - 2y)^3 + (2y - 4z)^3 + (4z - x)^3 = 3(x - 2y)(2y - 4z)(4z - x) \] --- ### Part (iii): Factorize \( (3\sqrt{2}a - 5\sqrt{3}b)^3 + (5\sqrt{3}b - 7\sqrt{5}c)^3 + (7\sqrt{5}c - 3\sqrt{2}a)^3 \) 1. **Identify \( a, b, c \)**: - Let \( a = 3\sqrt{2}a - 5\sqrt{3}b \) - Let \( b = 5\sqrt{3}b - 7\sqrt{5}c \) - Let \( c = 7\sqrt{5}c - 3\sqrt{2}a \) 2. **Check if \( a + b + c = 0 \)**: \[ a + b + c = (3\sqrt{2}a - 5\sqrt{3}b) + (5\sqrt{3}b - 7\sqrt{5}c) + (7\sqrt{5}c - 3\sqrt{2}a) = 0 \] The sum is zero, thus we can apply the identity. 3. **Apply the identity**: \[ (3\sqrt{2}a - 5\sqrt{3}b)^3 + (5\sqrt{3}b - 7\sqrt{5}c)^3 + (7\sqrt{5}c - 3\sqrt{2}a)^3 = 3(3\sqrt{2}a - 5\sqrt{3}b)(5\sqrt{3}b - 7\sqrt{5}c)(7\sqrt{5}c - 3\sqrt{2}a) \] ### Final Answer for Part (iii): \[ (3\sqrt{2}a - 5\sqrt{3}b)^3 + (5\sqrt{3}b - 7\sqrt{5}c)^3 + (7\sqrt{5}c - 3\sqrt{2}a)^3 = 3(3\sqrt{2}a - 5\sqrt{3}b)(5\sqrt{3}b - 7\sqrt{5}c)(7\sqrt{5}c - 3\sqrt{2}a) \] ---
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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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