Tag: unchanging relations

Quizzes (2) Questions (72)

Questions Related to unchanging relations

Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

If $a+b+c=0$ then $a^3+b^3+c^3$ is equal to

  1. 3abc
  2. $\displaystyle\frac{3}{abc}$
  3. $3a^3b^3c^3$

  4. zero
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Using\quad { a }^{ 3 }+\quad { b }^{ 3 }+\quad { c }^{ 3 }-3abc=\left( a+b+c \right) \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca \right) $

$Using\quad { a }^{ 3 }+\quad { b }^{ 3 }+\quad { c }^{ 3 }-3abc=0\times \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca \right) $
$Using\quad { a }^{ 3 }+\quad { b }^{ 3 }+\quad { c }^{ 3 }=3abc$
Here if a+b+c is 0 then answer will be 3abc


Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

If $\displaystyle a-\frac{1}{3}=\frac{1}{a}$ then the value of $\displaystyle a^{3}-\frac{1}{a^{3}}$ is

  1. $\displaystyle 1\frac{1}{27}$
  2. $\displaystyle 1\frac{2}{27}$
  3. $\displaystyle 1\frac{3}{27}$
  4. $\displaystyle 1\frac{4}{27}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $a-\frac{1}{3}=\frac{1}{a}$

=$a-\frac{1}{a}=\frac{1}{3}$
We know that
$x^{3}-y^{3}=(x-y)^{3}+3xy(x+y)$
Then $a^{3}-\left ( \frac{1}{a} \right )^{3}=(a-\left ( \frac{1}{a} \right ))^{3}+3a\left ( \frac{1}{a} \right )(a+\left ( \frac{1}{a} \right ))$ 
=$\left ( \frac{1}{3} \right )^{3}+3\times\left (  \frac{1}{3} \right )$
=$\frac{1}{27}+1=\frac{28}{27}=1\frac{1}{27}$