Without actually calculating the cubes- the value of -left - dfrac-1-2- right -3-left - dfrac-1-3- right -3-left - dfrac-5-6- right -3-dfrac-5-12-dfrac-5-12-dfrac-7-12-dfrac-7-12-

Question

Toppr Admin · Accepted Answer

We know that-a3-b3-c3-a-b-c-a2-b2-c2-x2212-ab-x2212-bc-x2212-ca-3abcIf a-b-c-0- then-xA0-a3-b3-c3-3abc

Without actually calculating the cubes, find the value of :
${(\frac{1}{2})}^{3} + {(\frac{1}{3})}^{3} - {(\frac{5}{6})}^{3}$

$\frac{5}{12}$
$- \frac{5}{12}$
$\frac{7}{12}$
$- \frac{7}{12}$

We know that,
$a^{3} + b^{3} + c^{3} = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) + 3 a b c$
If $a + b + c = 0$ , then $a^{3} + b^{3} + c^{3} = 3 a b c$

Without actually calculating the cubes, find the value of :(12)3+(13)3−(56)3512−512712−712

We know that,a3+b3+c3=(a+b+c)(a2+b2+c2−ab−bc−ca)+3abcIf a+b+c=0, then a3+b3+c3=3abc

Without actually calculating the cubes, find the value of :
${(\frac{1}{2})}^{3} + {(\frac{1}{3})}^{3} - {(\frac{5}{6})}^{3}$

$\frac{5}{12}$
$- \frac{5}{12}$
$\frac{7}{12}$
$- \frac{7}{12}$

We know that,
$a^{3} + b^{3} + c^{3} = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) + 3 a b c$
If $a + b + c = 0$ , then $a^{3} + b^{3} + c^{3} = 3 a b c$