Question

Q. If X=1/3-√8 , find the value of X cube - 2X square -7X +5​

Answer 1

ANSWER:

Given:

• x = 1/(3-√8)

To Find:

• Value of x³ - 2x² - 7x + 5

Solution:

$\text{We are given that,}\\\\:\longrightarrow x=\dfrac{1}{3-\sqrt{8}}\\\\\text{On rationalizing,}\\\\:\implies x=\dfrac{1}{3-\sqrt{8}}\times\dfrac{3+\sqrt{8}}{3+\sqrt{8}}\\\\:\implies x=\dfrac{1\times(3+\sqrt{8})}{(3-\sqrt{8})(3+\sqrt{8})}\\\\\text{We know that,}\\\\:\hookrightarrow(a+b)(a-b)=a^2-b^2\\\\:\implies x=\dfrac{3+\sqrt{8}}{3^2-(\sqrt{8})^2}\\\\:\implies x=\dfrac{3+\sqrt{8}}{9-8}\\\\:\implies x=3+\sqrt{8}- - - -(1)\\\\\text{We have to find value of:}\\\\:\longrightarrow x^3-2x^2-7x+5\\\\\text{Putting value of x from (1),}\\\\:\implies(3+\sqrt{8})^3-2(3+\sqrt{8})^2-7(3+\sqrt{8})+5$

$\text{We know that,}\\\\:\hookrightarrow(a+b)^3=a^3+b^3+3a^2b+3ab^2,\\\\:\hookrightarrow(a+b)^2=a^2+b^2+2ab\\\\\text{So,}\\\\:\implies(3+\sqrt{8})^3-2(3+\sqrt{8})^2-7(3+\sqrt{8})+5\\\\:\implies[3^3+(\sqrt{8})^3+3(3)^2(\sqrt{8})+3(3)(\sqrt{8})^2]-2[3^2+(\sqrt{8})^2+2(3)(\sqrt{8})]-7(3+\sqrt{8})+5\\\\:\implies[27+8\sqrt{8}+3(9)(\sqrt{8})+3(3)(8)]-2[9+8+6\sqrt{8}]-7(3+\sqrt{8})+5\\\\:\implies(27+8\sqrt{8}+27\sqrt{8}+72)-2(17+6\sqrt{8})-21-7\sqrt{8}+5\\\\:\implies99+35\sqrt{8}-34-12\sqrt{8}-21-7\sqrt{8}+5\\\\\text{Grouping like terms together,}\\\\:\implies99-34-21+5+35\sqrt{8}-12\sqrt{8}-7\sqrt{8}\\\\\text{Solving the like terms,}\\\\\bf{:\implies49+16\sqrt{8}}$

Formula Used:

• (a+b)(a-b) = a² - b²
• (a+b)³ = a³ + b³ + 3a²b + 3ab²
• (a+b)² = a² + b² + 2ab

Answer 2

best answer ke liye refer the attachment.

Agar achha lage to like zarur krna.