d Problem: factorise 2 root2 x^3 +3 root3y^3​ | Job Exam Rare Mathematics (JEMrare) - Solved MCQs
Advertisements

Problem: factorise 2 root2 x^3 +3 root3y^3​

Problem: factorise 2 root2 x^3 +3 root3y^3​

Solution:

The expression can be written as,

\[ 2\sqrt2x^3+3\sqrt3y^3 \]

This can be written as ,

\[ =2^1\times 2^{\frac{1}{2}}x^3 +3\times 3^{\frac{1}{2}}y^3 \]

or

\[ =2^{1+\frac{1}{2}}x^3+3^{1+\frac{1}{2}}y^3 \]

or

\[ =2^{\frac{3}{2}}x^3+3^{\frac{3}{2}}y^3 \]

Now to factorize it we will give it shape of ​\( a^3+b^3 \)​ formula !

So,

\[ = (2^\frac{1}{2}x)^3+(3^\frac{1}{2}y)^3 \]

So we know the formula;

\[ a^3+b^3=(a+b)(a^2-ab+b^2) \]

so,

\[ (2^\frac{1}{2}x)^3+(3^\frac{1}{2}y)^3=(2^\frac{1}{2}x+3^\frac{1}{2}y){(2^\frac{1}{2}x)^2}-2^\frac{1}{2}x3^\frac{1}{2}y+{(3^\frac{1}{2}y)^2} \]

Which can be further simplified as;

\[ (2^\frac{1}{2}x)^3+(3^\frac{1}{2}y)^3=(2^\frac{1}{2}x+3^\frac{1}{2}y)(2x^2-6^\frac{1}{2}xy+3y^2) \]

or

\[ (2^\frac{1}{2}x)^3+(3^\frac{1}{2}y)^3= (\sqrt2x+\sqrt3y)(2x^2-\sqrt6xy+\3y^2) \]

You may submit a MCQ or maths problem with solution tips here in the comment box below! If you could not solve anyone, you may submit a MCQ or maths problem without solution as well. We will try our level best to solve it for you!

Your email address will not be published.

You may use these <abbr title="HyperText Markup Language">HTML</abbr> tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <s> <strike> <strong>