Math, asked by anushree291207, 1 month ago

(√3-√2)^3+(√3+√2)^3 please solve this question​

Answers

Answered by sehgalp381
28

Answer:

( \sqrt{3}  -  \sqrt{2} ) {}^{3}  + ( \sqrt{3} +  \sqrt{2}  ) {}^{3}  \\ ( \sqrt{3} ) {}^{3}  + ( \sqrt{2} ) {}^{3}  - 3 \times  \sqrt{3}  \times  \sqrt{2} ( \sqrt{3}  -  \sqrt{2} ) + ( \sqrt{3} ) {}^{3}  + ( \sqrt{2} ) {}^{3}  + 3 \times  \sqrt{3}  \times  \sqrt{2} ( \sqrt{3}  +  \sqrt{2} )

3 \sqrt{3}  + 2 \sqrt{2}  - 3 \sqrt{6} ( \sqrt{3}  -  \sqrt{2} ) + 3 \sqrt{3}  + 2 \sqrt{2}  + 3 \sqrt{6} ( \sqrt{3}  +  \sqrt{2} ) \\

3 \sqrt{3}  + 2 \sqrt{2}  - 3 \sqrt{6}  - 3 \sqrt{18}  - 3 \sqrt{12}  + 3 \sqrt{3}  + 2 \sqrt{2}  + 3 \sqrt{6}  + 3 \sqrt{18}  + 3 \sqrt{12}

3 \sqrt{3}  + 2 \sqrt{2}  + 3 \sqrt{3}  + 2 \sqrt{2 }  \\ 3 \sqrt{3}  + 3 \sqrt{3}  + 2 \sqrt{2 }  + 2 \sqrt{2}  \\ 6 \sqrt{3}  + 4 \sqrt{2}

Step-by-step explanation:

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Answered by sunayksingh
0

Answer:

Answer:

(3−2)3+(3+2)3(3)3+(2)3−3×3×2(3−2)+(3)3+(2)3+3×3×2(3+2)\begin{lgathered}( \sqrt{3} - \sqrt{2} ) {}^{3} + ( \sqrt{3} + \sqrt{2} ) {}^{3} \\ ( \sqrt{3} ) {}^{3} + ( \sqrt{2} ) {}^{3} - 3 \times \sqrt{3} \times \sqrt{2} ( \sqrt{3} - \sqrt{2} ) + ( \sqrt{3} ) {}^{3} + ( \sqrt{2} ) {}^{3} + 3 \times \sqrt{3} \times \sqrt{2} ( \sqrt{3} + \sqrt{2} )\end{lgathered}

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33+22−36(3−2)+33+22+36(3+2)\begin{lgathered}3 \sqrt{3} + 2 \sqrt{2} - 3 \sqrt{6} ( \sqrt{3} - \sqrt{2} ) + 3 \sqrt{3} + 2 \sqrt{2} + 3 \sqrt{6} ( \sqrt{3} + \sqrt{2} ) \\\end{lgathered}

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33+22−36−318−312+33+22+36+318+3123 \sqrt{3} + 2 \sqrt{2} - 3 \sqrt{6} - 3 \sqrt{18} - 3 \sqrt{12} + 3 \sqrt{3} + 2 \sqrt{2} + 3 \sqrt{6} + 3 \sqrt{18} + 3 \sqrt{12}3

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33+22+33+2233+33+22+2263+42\begin{lgathered}3 \sqrt{3} + 2 \sqrt{2} + 3 \sqrt{3} + 2 \sqrt{2 } \\ 3 \sqrt{3} + 3 \sqrt{3} + 2 \sqrt{2 } + 2 \sqrt{2} \\ 6 \sqrt{3} + 4 \sqrt{2}\end{lgathered}

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Step-by-step explanation:

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