Math, asked by swati460, 1 year ago

Please give answer of the above question

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attinderpaul55225: what is the answer

Answers

Answered by attinderpaul55225
1

we \: know \: that \\  {a}^{3}  +  {b}^{3}  +  {c}^{3}  - 3abc  = 0 \\  =  >  {a}^{3}  +  {b}^{3}  +  {c}^{3}  = 3abc \\ here \: we \: can \: compair \\  {( {a}^{2}   -  {b}^{2} )}^{3}  =  {a}^{3}  \\  { ({b}^{2} -  {c}^{2} ) }^{3}  =  {b}^{3}  \\  {( {c}^{2} -  {a}^{2}  )}^{3}  =  {c}^{3}  \\ so \\  {( {a}^{2}   -  {b}^{2} )}^{3}  +  {( {b}^{2}  -  {c}^{2}) }^{3}  +  {( {c}^{2}  -  {a}^{2} )}^{3}  \\   \\ = 3.( {a}^{2}  -  {b}^{2} ).( {b}^{2}  -  {c}^{2} ).( {c}^{2}  -  {a}^{2} ) \\  \\  = 3.(a + b)(a - b)(b  + c)(b - c)(c + a)(c - a) \\  \\ next \\  \\  {(a - b)}^{3}  + {(b - c)}^{3}  +  {(c - a)}^{3}  \\  \\  = 3.(a - b).(b - c).(c - a) \\ now \:  \\  \\ \frac{ {( {a}^{2}  -  {b}^{2} ) }^{3} +  {( {b}^{2}  -  {c}^{2} ) }^{3}   +  {( {c}^{2}  -  {a}^{2} )}^{3} }{ {(a - b)}^{3}  +  {(b - c)}^{3} }  +  {(c - a)}^{3}  \\  \\  =  \frac{3(a + b)(a - b)(b + c)(b - c)(c + a)(c - a)}{3.(a - b)(b - c)(c - a)}  \\  = 3.(a + b)(b + c)(c - a)

the answer. is

3.(a+b)(b+c)(c+a) ✔✔✔✔✔✔


swati460: Thanks a lot
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