Question

If a, b and c are integers such that $(\sqrt[3]{34}+\sqrt[3]{32}-2)(\sqrt[3]{34a}+\sqrt[3]{34a}+\sqrt[3]{32b}+c)=20$, then a + b – c equals

• A. 8
• B. 10
• C. 20
• D. 5

Answer 1

The correct option is B 10

$(\sqrt[3]{34}+\sqrt[3]{32}-2)(\sqrt[3]{34a}+\sqrt[3]{34a}+\sqrt[3]{32b}+c)=20$

$\Rightarrow (2^{\frac{2}{3}}+2^{\frac{1}{3}}-2)(2^{\frac{2}{3}}a+2^{\frac{1}{3}}b+c)=20$

$\Rightarrow 2^{\frac{2}{3}}(2^{\frac{2}{3}}a+2^{\frac{1}{3}}b+c)+2^{\frac{1}{3}}(2^{\frac{2}{3}}a+2^{\frac{1}{3}}b+c)-2(2^{\frac{2}{3}}a+2^{\frac{1}{3}}b+c)=20$
$\Rightarrow 2^{\frac{4}{3}}a+2^{\frac{3}{3}}b+2^{\frac{2}{3}}c+2^{\frac{3}{3}}a+2^{\frac{2}{3}}b+2^{\frac{1}{3}}c-2^{\frac{5}{3}}a-2^{\frac{4}{3}}b-2c=20$
$\Rightarrow (-2^{\frac{5}{3}}a+2^{\frac{2}{3}}b+2^{\frac{2}{3}}c)+(2^{\frac{4}{3}}a-2^{\frac{4}{3}}b+2^{\frac{1}{3}}c)+(2a+2b-2c)=20$
$\Rightarrow 2^{\frac{2}{3}}(-2a+b+c)+2^{\frac{1}{3}}(2a-2b+c)+(2a+2b-2c)=20$
On comparing the left side and right of the equation: –2a + b + c = 0, 2a – 2b + c = 0 and
2a + 2b – 2c = 20
⇒ 2(a + b – c) = 20
⇒ a + b – c = 10
Hence, the correct answer is option (b).