Question

1) (a+b)^ 3 = a^3 + 3a^2b + 3ab^2 + b^3. 2)(a+b)^3 = a^3 + b^3 + 3ab(a+b). 3)(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3.
4)(a-b)^3 = a^3 - b^3 -
3ab(a-b). with the help of the said formulas , solve the following questions:. (1) if (2x + 1/3x) = 4, try to prove: 27x^3 + 1/8x^3 = 189. (2)2a - 2/a + 1 = 0, calculate the value of a^3 - 1/a^3 +2
(ans):3/8.
(3) if a^3 + b^3 + c^3 = 3abc , find out the value of (a+b+c) [where a is not equal to b and b is not equal to c.
(ans:0) (4) if m+n = 5 and mn=6, find out the value of. (m^2 + n^2)(m^3 + n^3).
(ans:455) ​

Answer 1

kindly check the attachment also

3)

IDENTITY USED:-

${a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - ac - bc)$

GIVEN:-

${a}^{3} + {b}^{3} + {c}^{3} = 3abc$

so,

${a}^{3} + {b}^{3} + {c}^{3} - 3abc = 0$

$(a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - ac - bc) = this \: either \: means( {a}^{2} + {b}^{2} + {c}^{2} - ab - ac - bc) = 0 \: or \: (a + b + c) = 0$

$( {a}^{2} + {b}^{2} + {c}^{2} - ab - ac - bc) = 0 \: cannot \: be \: zero \: bcz \:$

$2 {a}^{2} + 2 {b}^{2} + 2 {c}^{2} - 2ab - 2ac - 2bc = 0$

${a}^{2} + {b}^{2} - 2ab + {a}^{2} + {b}^{2} + 2 {c}^{2} - 2ac - 2bc = 0$

$(a - b {)}^{2} + {a}^{2} + {c}^{2} - 2ac + {b}^{2} + {c }^{2} - 2bc = 0$

$(a - b {)}^{2} + (a - c {)}^{2} + (b - c {)}^{2} = 0$

$(a - {b)}^{2} ,(a - {c)}^{2} ,(b - {c)}^{2} > = 0$