Question

if (4+3/x)(16-12/x+9/x^2)=5 then find (4-3/x)(16+12/x+9/x^2)

Answer 1

Answer:

$=\frac{1}{5} [16^3 - (\frac{9}{x^2})^3]$

Step-by-step explanation:

$(4 + \frac{3}{x} )(16 - \frac{12}{x} + \frac{9}{x^{2} } ) = 5$ [let's call this as equation no. i]

Let's assume $(4 - \frac{3}{x} )(16 + \frac{12}{x} + \frac{9}{x^{2} } ) = y$  [let's call this as equation no. ii]

Now multiplying equation i by equation ii:

$(4 + \frac{3}{x} )(4 - \frac{3}{x} )(16 - \frac{12}{x} + \frac{9}{x^{2} } )(16 + \frac{12}{x} + \frac{9}{x^{2} } ) = 5y$

5y $=[4^{2} - (\frac{3}{x}) ^{2} ][(16+\frac{9}{x^{2} }) ^{2} - (\frac{12}{x} )^2]$

$=[16 - (\frac{9}{x^2}) ][(16^2+\frac{288}{x^{2} } + \frac{81}{x^4} ) - (\frac{12}{x} )^2]$

$=[16 - (\frac{9}{x^2}) ][(16^2+\frac{288}{x^{2} } + \frac{81}{x^4} ) - \frac{144}{x^2} ]$

$=[16 - (\frac{9}{x^2}) ][(16^2-\frac{144}{x^{2} } + \frac{81}{x^4} ) ]$

$=[16^3 - (\frac{9}{x^2})^3 ]$  [∵ $(a - b)(a^2 -ab + b^2) = a^3 - b^3$]

$=16^3 - (\frac{9}{x^2})^3$

∴ y $=\frac{1}{5} [16^3 - (\frac{9}{x^2})^3]$