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without actually calculating , the cubes , find the valuse of
(i) `((1)/(2))^(3)+((1)/(3))^(3)-((5)/(6))^(3)`(ii) `(0.2)^(3)-(0.3)^(3)+(0.1)^(3)`

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To solve the given problems without actually calculating the cubes, we can use the identity for the sum of cubes. The identity states that if \( a + b + c = 0 \), then: \[ a^3 + b^3 + c^3 = 3abc \] Let's apply this identity to both parts of the question. ...
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Without actually calculating the cube: (-12)^3+7^3+5^3

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NCERT EXEMPLAR ENGLISH-POLYNOMIALS-EXERCISE 2.2 very short Answer type Questions
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  3. Factorise (i) x^(2)+9x+18 (ii) 6x^(2)+7x-3 (iii) 2x^(2)-7x-15 (...

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  8. Expand the following (i) (4a-b+ac)^(2) (ii) (3a-5b-c)^(2) (iii)...

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  9. Facorise the following (i) 9x^(2)+4y^(2)+16z^(2)+12xy-16yz-24xz (...

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  10. if a +b+c=9 and ab+bc+ca=26,find a^(2)+b^(2)+c^(2).

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  11. Expand the following (i) (3a-2b)^(3) (ii) ((1)/(x)+(y)/(3))^(3) (ii...

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  12. Facorise the following (i) 1-64a^(3)-12+48a^(2) (ii) 8p^(3)+(12)...

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  13. find the following products (i) ((x)/(2)+2y)((x^(2))/(4)-xy+4y^(2...

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  14. Factorise (i) 1+64x^(3)(ii) a^(3)-2sqrt(2)b^(3)

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  15. Find the product: (2x-y+3z)(4x^2+y^2+9z^2+2x y+3y z-6x z)

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  16. Factorise (i) a^(3)+8b^(3)+64c^(3)-24abc

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  17. without actually calculating , the cubes , find the valuse of ...

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  18. Factorize : (x-2y)^3+(2y-3z)^3+(3z-x)^3

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  19. Find the value of : (1) x^3 + y^3 -12xy + 64 when x+y+4=0 (2) x^3 ...

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  20. Give possible expression for the length and breadth of the rec...

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