Question

if a is equal to under root 7 + 4 root 3 upon 7 minus 4 root 3 then the value of a bracket a minus 14 the whole raise to 2 is​

Answer 1

Answer:

The required value is (x^3+\frac{1}{x^3})=2702(x3+x31)=2702

Step-by-step explanation:

Given : x=7+4\sqrt3x=7+43

To find : (x^3+\frac{1}{x^3})(x3+x31)

Solution :

We know, x=7+4\sqrt3x=7+43 .....(1)

\frac{1}{x}=\frac{1}{7+4\sqrt3}x1=7+431

Rationalizing,

\frac{1}{x}=\frac{1}{7+4\sqrt3}\times \frac{7-4\sqrt3}{7-4\sqrt3}x1=7+431×7−437−43

\frac{1}{x}=\frac{7-4\sqrt3}{7^2-(4\sqrt3)^2}x1=72−(43)27−43

\frac{1}{x}=\frac{7-4\sqrt3}{49-48}x1=49−487−43

\frac{1}{x}=7-4\sqrt3x1=7−43 ....(2)

Cubing equation (1),

x^3=(7+4\sqrt3)^3x3=(7+43)3

x^3=(7)^3+(4\sqrt3)^3+3(7)^2(4\sqrt3)+3(7)(4\sqrt3)^2x3=(7)3+(43)3+3(7)2(43)+3(7)(43)2

x^3=343+192\sqrt3+588\sqrt3+1008x3=343+1923+5883+1008

x^3=1351+780\sqrt3x3=1351+7803 ......(3)

Cubing equation (2),

\frac{1}{x^3}=(7-4\sqrt3)^3x31=(7−43)3

\frac{1}{x^3}=(7)^3+(-4\sqrt3)^3+3(7)^2(-4\sqrt3)+3(7)(-4\sqrt3)^2x31=(7)3+(−43)3+3(7)2(−43)+3(7)(−43)2

\frac{1}{x^3}=343-192\sqrt3-588\sqrt3+1008x31=

Answer 2

Answer:

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