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x+12x=3x + \frac{1}{2x} = 3 ആണെങ്കിൽ 8x3+1x38x^3+ \frac{1}{x^3} കാണുക

A213

B216

C180

Dnone of the above

Answer:

C. 180

Read Explanation:

നൽകിയത്:

x+12x=3x + \frac{1}{2x} = 3

ആദ്യം 2 കൊണ്ട് multiply ചെയ്യാം:

2x+1x=62x + \frac{1}{x} = 6

Identity ഉപയോഗിക്കാം:

(a+b)3=a3+b3+3ab(a+b)(a+b)^3 = a^3 + b^3 + 3ab(a+b)

ഇവിടെ (a = 2x), (b = \frac{1}{x})


(2x+1x)3=(2x)3+(1x)3+3(2x1x)(2x+1x)(2x + \frac{1}{x})^3 = (2x)^3 + \left(\frac{1}{x}\right)^3 + 3(2x \cdot \frac{1}{x})(2x + \frac{1}{x})

Substitute ചെയ്യാം:

(2x+1x)3=8x3+1x3+32(2x+1x)(2x + \frac{1}{x})^3 = 8x^3 + \frac{1}{x^3} + 3 \cdot 2 \cdot (2x + \frac{1}{x})

ഇപ്പോൾ value substitute ചെയ്യുക:

63=8x3+1x3+6(6)6^3 = 8x^3 + \frac{1}{x^3} + 6(6)

216=8x3+1x3+36216 = 8x^3 + \frac{1}{x^3} + 36

Final answer:

8x3+1x3=21636=1808x^3 + \frac{1}{x^3} = 216 - 36 = \boxed{180}


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