Question

24. If a^2 + b^2+ c^2 + 27 = 6 (a + b + c), then what is the value of
$\sqrt[3]{a ^{3} } + b ^{3} + c ^{3}$
​

Answer 1

ANSWER:

Given:

• $a^2+b^2+c^2+27=6(a+b+c)$

To Find:

• $\text{Value of: }\sqrt[3]{a^3+b^3+c^3}$

Solution:

$\text{We are given that,}\\\\:\longrightarrow a^2+b^2+c^2+27=6(a+b+c)\\\\\text{Solving the equation,}\\\\:\implies a^2+b^2+c^2+27=6a+6b+6c\\\\\text{Transposing RHS to LHS,}\\\\:\implies a^2+b^2+c^2+27-6a-6b-6c=0\\\\\text{Adding and subtracting 9 three times,}\\\\:\implies a^2+b^2+c^2+27-6a-6b-6c+9+9+9-9-9-9=0\\\\\text{Rearranging the terms,}\\\\:\implies a^2-6a+9+b^2-6b+9+c^2-6c+9+27\!\!\!\!/\,-27\!\!\!\!/\,=0\\\\\text{Grouping the terms,}\\\\:\implies(a^2-6a+9)+(b^2-6b+9)+(c^2-6c+9)=0\\\\:\implies[a^2-2(a)(3)+3^2]+[b^2+2(b)(3)+3^2]+[c^2-2(c)(3)+3^2]=0\\\\\text{We know that,}\\\\:\hookrightarrow x^2-2xy+y^2=(x-y)^2\\\\\text{So,}\\\\:\implies(a-3)^2+(b-3)^2+(c-3)^2=0$

$\text{For the given equation to be true,}\\\\\text{All 3 terms should be 0.}\\\\\text{So,}\\\\:\longrightarrow(a-3)^2=0 \implies a-3=0\\\\:\implies a=3- - - -(1)\\\\:\longrightarrow(b-3)^2=0 \implies b-3=0\\\\:\implies b=3- - - -(2)\\\\:\longrightarrow(c-3)^2=0 \implies c-3=0\\\\:\implies c=3- - - -(3)\\\\\text{We need to find,}\\\\:\longrightarrow\sqrt[3]{a^3+b^3+c^3}\\\\\text{From (1), (2) & (3),}\\\\:\implies\sqrt[3]{(3)^3+(3)^3+(3)^3}\\\\:\implies\sqrt[3]{27+27+27}\\\\:\implies\sqrt[3]{81}\\\\:\implies\sqrt[3]{3\times3\times3\times3}\\\\\bf{:\implies3}\sqrt[\bf{3}]{\bf{3}}$

Answer 2

best answer ke liye refer the attachment.

Agar achha lage to like zarur krna.