f(x)=(2x-x^2)^3 eching

$$f ( x ) = ( 2 x - x ^ { 2 } ) ^ { 3 }$$

Baholash

$$\left(x\left(2-x\right)\right)^{3}$$

Yechim qadamlari

\left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} binom teoremasini \left(2x-x^{2}\right)^{3} kengaytirilishi uchun ishlating.

$$8x^{3}-12x^{2}x^{2}+6x\left(x^{2}\right)^{2}-\left(x^{2}\right)^{3}$$

Ayni asosning daraja ko‘rsatkichlarini ko‘paytirish uchun ularning darajalarini qo‘shing. 2 va 2 ni qo‘shib, 4 ni oling.

$$8x^{3}-12x^{4}+6x\left(x^{2}\right)^{2}-\left(x^{2}\right)^{3}$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va 2 ni ko‘paytirib, 4 ni oling.

$$8x^{3}-12x^{4}+6xx^{4}-\left(x^{2}\right)^{3}$$

Ayni asosning daraja ko‘rsatkichlarini ko‘paytirish uchun ularning darajalarini qo‘shing. 1 va 4 ni qo‘shib, 5 ni oling.

$$8x^{3}-12x^{4}+6x^{5}-\left(x^{2}\right)^{3}$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va 3 ni ko‘paytirib, 6 ni oling.

$$8x^{3}-12x^{4}+6x^{5}-x^{6}$$

Kengaytirish

$$8x^{3}-12x^{4}+6x^{5}-x^{6}$$

Yechim qadamlari

\left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} binom teoremasini \left(2x-x^{2}\right)^{3} kengaytirilishi uchun ishlating.

$$8x^{3}-12x^{2}x^{2}+6x\left(x^{2}\right)^{2}-\left(x^{2}\right)^{3}$$

Ayni asosning daraja ko‘rsatkichlarini ko‘paytirish uchun ularning darajalarini qo‘shing. 2 va 2 ni qo‘shib, 4 ni oling.

$$8x^{3}-12x^{4}+6x\left(x^{2}\right)^{2}-\left(x^{2}\right)^{3}$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va 2 ni ko‘paytirib, 4 ni oling.

$$8x^{3}-12x^{4}+6xx^{4}-\left(x^{2}\right)^{3}$$

Ayni asosning daraja ko‘rsatkichlarini ko‘paytirish uchun ularning darajalarini qo‘shing. 1 va 4 ni qo‘shib, 5 ni oling.

$$8x^{3}-12x^{4}+6x^{5}-\left(x^{2}\right)^{3}$$

Daraja ko‘rsatkichini boshqa ko‘rsatkichga oshirish uchun, darajalarini ko‘paytiring. 2 va 3 ni ko‘paytirib, 6 ni oling.

$$8x^{3}-12x^{4}+6x^{5}-x^{6}$$